Initial speed of ball will be 2.7m/s.
When a body is rotating about an axis, then it has kinetic energy.
And this energy is called rotational kinetic energy.
It is given as - R.K.E. = 1/2 Iω²
And if a ball is rolling without slipping.
Then the moment of inertia of the solid ball is written as -
I = 25MR²
Vi = Rω
Here it is given in the problem that-
height(h) = 0.53m
Now by the conservation of energy we can write the equation as -
1/2MVi² + 1/2Iω² = Mgh
so that -
(1/2)MVi² + (1/2)×(2/5MR²) ×(Vi/R)² = Mgh(1/2)Vi² + (1/5)Vi²
= gh(7/10)Vi² = 9.8 × 0.53
Vi = 2.7 m/s
So that the initial velocity of ball came out to be 2.7m/s after applying all concepts or rotational motion.
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The weather map shows some conditions in the atmosphere at noon on a
particular day.
Where would you expect the warmest weather?
A. In the southwest
B. In the northwest
C. In the southeast
D. In the northeast.
The geographical region where I would expect the warmest weather is: C. in the southeast.
What is a weather map?A weather map can be defined as a type of chart that is typically used to provide information about the average atmospheric condition of a particular geographical region over a specific period of time.
Based on the weather map shown in the image attached below, we can infer and logically deduce that the geographical region where the warmest weather is expected is in the southeast due to its very high atmospheric pressure.
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The initial velocity of a car moving along a straight path was 0 meters/second. After 10 seconds, it reached a velocity of 25 meters/second. Calculate the car’s average acceleration during this time period.
Answer:
[tex]2.5m/s^{2}[/tex]
Explanation:
We can use one of the kinematics equation to find the acceleration.
v = u + at
where v = final velocity, in m/s
u = initial velocity, in m/s
t = time in seconds
Given the information provided from the question,
v = 25m/s
u = 0m/s
t = 10s
Substitute these values into the kinematics equation.
25 = 0 + 10a
10a = 25
a = 25 / 10
= [tex]2.5m/s^{2}[/tex]
Shown below is a box of mass m1 that sits on a frictionless incline at an angle above the horizontal θ=30°. This box is connected by a relatively massless string, over a frictionless pulley, and finally connected to a box at rest over the ledge, labeled m2. If m1 and m2 are a height h above the ground and m2>>m1: (a) What is the initial gravitational potential energy of the system? (b) What is the final kinetic energy of the system?
(a) The initial gravitational potential energy of the system is 0.5 (m₁ + m₂)gL.
(b) The final kinetic energy of the system is 0.5 (m₁ + m₂)gL .
Initial gravitational potential energy of the system
P.E = mgh
where;
h is height of the inclineLet the length of the incline = L
h = L sinθ
P.E = mg(L sinθ)
P.E = (m₁ + m₂)g (L sinθ)
P.E = 0.5 (m₁ + m₂)gL
Final kinetic energy of the systemThe final kinetic energy of the system is equal to the initial potential energy since the inclined plane is frictionless.
K.E(f) = (m₁ + m₂)g (L sin30)
K.E(f) = 0.5 (m₁ + m₂)gL
Thus, the initial gravitational potential energy of the system is 0.5 (m₁ + m₂)gL and the final kinetic energy of the system is0.5 (m₁ + m₂)gL .
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Sally and Sam are in a spaceship that comes to within 14,000 km of the asteroid Ceres. Determine the force Sally experiences, in N, due to the presence of the asteroid. The mass of the asteroid is 8.7 1020 kg and the mass of Sally is 67 kg. For calculation purposes, assume the two objects to be point masses.
Sam and Sally are traveling aboard a spacecraft that approaches the asteroid Ceres within 14,000 kilometers. Sally will experience 1.989 × 10⁻¹¹ N of force.
What is the gravitational force?Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.
The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.
Given data
Mass of asteroid ,m₁ = 8.7 1020 kg
Mass of sally,m₂ = 67 kg
Gravitational constant,G = 6.6 × 10⁻¹¹ kg⁻² m²
Distance of seperation,R = 14,000 km
[tex]\rm F = G\frac{m_1m_2}{R^ 2} \\\\ F = 6.6 \times 10^{-11 }\times \frac{8.71020 \times 67 }{(14,000)^2} \\\\F = 1.989 \times 10^{-11 } \ N[/tex]
Hence, the force Sally experiences will be 1.989 × 10⁻¹¹ N.
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Ten bricks, each 6.0 cm thick and mass 1.5 kg, lie flat on a table. How much work is required to stack them over eachother
The workdone required to stack them over each other is 39.69 J
What is workdone?This is defined as the product of force and distance moved in the direction of the force.
Workdone (Wd) = Force × distance (d)
But
Force = weight = mass × acceleration due to gravity
w = mg
Distance (d) = height (h)
Thus,
Wd = weight × height
How to determine the workdone in stacking them togetherWe'll begin by calculating the weight of each bricks
Mass of each bricks (m) = 1.5 KgAcceleration due to gravity (g) = 9.8 m/s²Weight of each bricks (W) =?W = mg
W = 1.5 × 9.8
W = 14.7 N
Next, we shall determine the work done in stacking the 2nd on the 1st brick and so on..
Workdone in stacking the 2nd on the 1st
Weight (W) = 14.7 NHeight (h) = 6 cm = 6 / 100 = 0.06 mWorkdone (Wd) = ?Wd = Weight × height
Wd = 14.7 × 0.06
Wd = 0.882 J
Workdone in stacking the 3rd
Weight (W) = 14.7 NHeight (h) = 0.12 mWorkdone (Wd) = ?Wd = Weight × height
Wd = 14.7 × 0.12
Wd = 1.764 J
Workdone in stacking the 4th
Weight (W) = 14.7 NHeight (h) = 0.18 mWorkdone (Wd) = ?Wd = Weight × height
Wd = 14.7 × 0.18
Wd = 2.646 J
Workdone in stacking the 5th
Weight (W) = 14.7 NHeight (h) = 0.24 mWorkdone (Wd) = ?Wd = Weight × height
Wd = 14.7 × 0.24
Wd = 3.528 J
Workdone in stacking the 6th
Weight (W) = 14.7 NHeight (h) = 0.3 mWorkdone (Wd) = ?Wd = Weight × height
Wd = 14.7 × 0.3
Wd = 4.41 J
Workdone in stacking the 7th
Weight (W) = 14.7 NHeight (h) = 0.36 mWorkdone (Wd) = ?Wd = Weight × height
Wd = 14.7 × 0.36
Wd = 5.292 J
Workdone in stacking the 8th
Weight (W) = 14.7 NHeight (h) = 0.42 mWorkdone (Wd) = ?Wd = Weight × height
Wd = 14.7 × 0.42
Wd = 6.174 J
Workdone in stacking the 9th
Weight (W) = 14.7 NHeight (h) = 0.48 mWorkdone (Wd) = ?Wd = Weight × height
Wd = 14.7 × 0.48
Wd = 7.056 J
Workdone in stacking the 10th
Weight (W) = 14.7 NHeight (h) = 0.54 mWorkdone (Wd) = ?Wd = Weight × height
Wd = 14.7 × 0.54
Wd = 7.938 J
Total workdone = 0.882 + 1.764 + 2.646 + 3.528 + 4.41 + 5.292 + 6.174 + 7.056 + 7.938
Total workdone = 39.69 J
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The WEIGHT of each brick is (m g) = 14.7 Newtons
Work to lift each brick = (force x distance) = [14.7 x (distance lifted)] Joules
Place 2nd brick on top of the 1st: lift 6cm, (14.7N x .06m) = 0.882 J
Place 3rd brick on top of the 2nd: lift 12cm, (14.7N x .12m) = 1.764 J
Place 4th brick on top of the 3rd: lift 18cm, (14.7N x .18m) = 2.646 J
Place 5th brick on top of the 4th: lift 24cm, (14.7N x .24m) = 3.528 J
Place 6th brick on top of the 5th: lift 30cm, (14.7N x .3m) = 4.41 J
Place 7th brick on top of the 6th: lift 36cm, (14.7N x .36m) = 5.292 J
Place 8th brick on top of the 7th: lift 42cm, (14.7N x .42m) = 6.174 J
Place 9th brick on top of the 8th: lift 48cm, (14.7N x .48m) = 7.056 J
Place 10th brick on top of the 9th: lift 54cm, (14.7N x .54m) = 7.938 J
Total work to stackum (addum up) = 39.7 Joules
____________________________________________
Way #2 (easier):
Weight of each brick = (mg) = 14.7 Newtons
Work to stack each one = (14.7N) x (distance lifted) Joules
Number of bricks to lift = 9
AVERAGE distance to lift them = (1/2) (54cm + 6cm) = 30cm
Average work = (weight of each brick) x (average distance lifted)
Total work = (number of bricks) x (weight of each) x (average lift)
Total work = (9 bricks) x (14.7N) x (0.3m) = 39.69 Joules
if a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the
direction 20° south of east, and finally walks 28 m in the direction 30° west of north.(2pt)
a) How far and at what angle is the Aster's final position from her initial position?
b) In what direction would she has to head to return to her initial position?
a. Aster is 56.3 m at 3.16° north-east from her initial position
b. She has to head to 183.16° or 86.84° south of west to return to her initial position
a. How to calculate how far Aster's final position from her initial position?Let Aster's initial position be represented by the vector r = 0i + 0j
Since she then walks walks first 70 m in the direction 37° north of east, let this displacement be represented by the vector u = (70sin37°)i + (70cos37°)j
= (70 0.6018)i + (70 0.7986)j
= 42.13i + 55.9j
Also, she then walks 82 m in the direction 20° south of east. Let this displacement be represented by the vector v = (82sin20°)i - (82cos20°)j
= -(82 0.3420)i + (82 0.9397)j
= 28.05i - 77.05j
Finally, she walks 28 m in the direction 30° west of north. Let this displacement be represented by the vector, w = -(28sin30°)i + (28cos30°)j = -(28 0.5)i + (28 0.8660)j
= -14i + 24.25j m
So, the total displacement is R = r + u + v + w
= 42.13i + 55.9j + 28.05i + (-77.05)j + (-14)i + 24.25j m
= 56.18i + 3.1j
So, how far she walks is the magnitude of R. The magnitude of a vector Z = xi + yj is Z = √(x² + y²)
So, the magnitude of R = √((56.18)² + (3.1)²)
= √(3156.19 + 9.61)
= √3165.8
= 56.3 m
Her direction from final position to initial positionThe direction of a vector Z = xi + yj is given by Ф = tan⁻¹ (y/x)
So, the direction of R is Ф' = tan⁻¹ (3.1/56.18)
= tan⁻¹ (0.0552)
= 3.16°
So, Aster is 56.3 m at 3.16° north-east from her initial position
b. What direction would she has to head to return to her initial position?To return to her original position, the displacement vector is V = r - R
= 0i + 0j - (56.18i + 3.1j)
= -56.18i - 3.1j
So, the direction of V is Ф" = tan⁻¹ (-3.1/-56.18)
= tan⁻¹ (0.0552)
= 3.16°
Since this is in the third quadrant, we have that the direction she must go to return to her original position is α = 180° + 3.16°
= 183.16°
or 90° - 3.16°
= 86.84° south of west
So, she has to head to 183.16° or 86.84° south of west to return to her initial position
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The larger the mass of a star, the higher the internal pressures. Higher internal pressures causes higher temperatures and it is temperature that determines the types of fusion that can occur deep in a stars interior. Discuss the types of fusion that can occur in a star, the temperatures at which they occur, and the mass required to produce them.
The types of fusion that occurs in stars are as follows:
The proton-proton fusion - stars with temperatures less than 15 million Kelvin, minimum masses of 0.08 solar mass.The carbon cycle fusion - stars with core temperatures greater than 15 million Kelvin, minimum masses of 4.0 solar mass.The helium fusion - stars with core temperatures greater than 100 million Kelvin; minimum masses of 0.5 solar mass.What is nuclear fusion?Nuclear fusion is the process by which nucleus of small atoms combine together to produces atoms of larger nucleus.
The types of nuclear fusion that can occur in a star are;
proton-proton fusion,helium fusion,the carbon cycle fusion.The proton-proton fusion occur in stars which have core temperatures less than 15 million Kelvin having minimum masses of 0.08 solar mass.
The carbon cycle fusion occurs in stars with core temperatures greater than 15 million Kelvin having minimum masses of 4.0 solar mass.
The helium fusion occurs in stars with core temperatures greater than 100 million Kelvin having minimum masses of 0.5 solar mass.
In conclusion, fusion reactions occurring in the stars produce the energy released by stars.
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DNA is unique to each __________.
Answer:
ORGANISM
Explanation:
Near the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of 3.45 m/s, and the second a velocity of 4.25 m/s.
(a) What is the velocity of the second runner relative to the first?
m/s faster than the front runner.
(b) If the front runner is 250 m from the finish line, who will win the race, assuming they run at constant velocity?
The first runner will win.
The second runner will win.
(c) What distance ahead will the winner be when she crosses the finish line?
m
a)The velocity of the second runner relative to the first will be 0.8 m/s.
b)The second runner will win.
c) The distance ahead will the winner be when she crosses the finish line will be 55.6 m.
What is velocity?The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
Given data;
The relative distance between the runner, S = 45.0 m.
The velocity of the front runner,v₁ = 3.45 m/s
The velocity of the second runner,v₂ = 4.25 m/s,
The distance of front runners from the finish line =250 m
The velocity of the second runner relative to the first is,v₂₁=?
The first runner's time to cross the finish line is,t₁
The second runner's time to cross the finish line is recorded,t₂
The distance ahead will the winner be when she crosses the finish line is, S₂₁
a)
The velocity of the second runner relative to the first is found as;
v₂₁ = v₂-v₁
v₂₁ =4.25-3.45
v₂₁ = 0.8 m/s in the runners' direction of travel.
b)
The first runner's time to cross the finish line is;
t₁ = 250/3.45
t₁=72.5 secs.
The second runner's time to cross the finish line is recorded;
t₂=(250+45)/4.25
t₂=69.4 secs.
So, the winner is the second runner.
c)
The winner will be in front when she crosses the finish line by a distance, The second runner travels a distance of within this time.
S₂ = 250+45
S₂=295 m.
The distance ahead will the winner be when she crosses the finish line is;
S₂₁ = 295-239.4
S₂₁=55.6
Hence, the velocity of the second runner relative to the first will be 0.8 m/s, the second runner will win. and the distance ahead will the winner be when she crosses the finish line will be 55.6 m.
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A boat moves through the water with two forces acting on it. One is a 1,800-N forward push by the water on the propeller, and the other is a 1,200-N resistive force due to the water around the bow.
(a) What is the acceleration of the 1,400-kg boat?
0.25
m/s 2
(b) If it starts from rest, how far will the boat move in 20.0 s?
m
(c) What will its velocity be at the end of that time?
m/s
(a) The 1,400-kg boat will accelerate at 0.425 m/sec².
(b) If the boat starts off at rest, it will travel 85 meters in 20.0 seconds.
(c)At the conclusion of the period, the speed will be 8.5 m/sec.
What is velocity?The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
Given data;
Forwad force acting on the boat,v₁ = 1,800-N
Resistive force acting on the boat,v₂ = 1,200-N
The acceleration of the boat,a
Mass of boat,m = 1,400-kg
Initial velocity of boat,u= 0 m/sec
Distance travelled by boat,S = ?
Time for the boat travels,t = 20.0 s
Final velocity,V = ? m/sec
The net force on the boat;
F = F₁ - F₂
F = 1800 N - 1200 N
F = 600 N
From the defination of force;
F= ma
a = F / m
a = 600 N / 1400 kg
a = 0.425 m/sec²
b)
The distance through which the boat moves is 20.0 s;
[tex]\rm x_f = x_ 0 + v_0 t + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}at^2 \\\\ x_f = 0+0 + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}\times 0.425 \times (20 )^ 2 \\\\ x_f = 85 \ m[/tex]
c)
The velocity at the end of that time is found as;
[tex]\rm v_f = v_ 0 + at \\\\ v_f =- 0+ at \\\\ v_f = 8.5[/tex] m/sec
Therefore, the boat's acceleration, the distance it travels in 20.0 seconds, and its final speed will be 0.425 m/sec², 85 m, and 8.5 m/sec.
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A car with a mass of 1080 kg is traveling in a mountainous area with a constant speed of 69.6 km/h. The road is horizontal and flat at point A, horizontal and curved at points B and C.
1. The radii of curvatures at B and C are: rB = 130 m and rC = 115 m. Calculate the normal force exerted by the road on the car at point A.
Normal Force = 1.-6x10^4 N
2. Now calculate the normal force exerted by the road on the car at point B.
(1) The normal force exerted by the road on the car at point A is 1.06 x 10⁴ N
(2) The normal force exerted by the road on the car at point B is 7,495.8 N.
Normal force at point AFn(A) = mg
where;
m is mass of the carg is acceleration due to gravityFn(A) = 1080 x 9.81
Fn(A) = 1.06 x 10⁴ N
Normal force exerted by the road on the car at point BThe normal force exerted by the road on the car at point B is calculated as follows;
Fn = Fn(A) - mv²/r
where;
Fn(A) is normal force at point Am is mass of the carr is radius of curve Bv is velocity of the car, 69.6 km/h = 19.33 m/sFn = 1.06 x 10⁴ - (1080 x 19.33²)/(130)
Fn = 7,495.8 N
Thus, the normal force exerted by the road on the car at point A is 1.06 x 10⁴ N and the normal force exerted by the road on the car at point B is 7,495.8 N.
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A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and 272.0 N (54.4 lbs) stand on the board so that the board is balanced.
What is the upward force exerted on the board by the support?
The upward force exerted on the board by the support is 530.8 N.
Upward force exerted on the board by the supportThe sum of the upward forces is equal to sum of downward forces;
total downward forces = 52.8 N + 206 N + 272 N = 530.8 N
downward force = upward force = 530.8 N
Thus, the upward force exerted on the board by the support is 530.8 N.
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The speed of supernova debris – about 15 years after Supernova 1987A exploded, its wind was seen hitting a pre-existing ring of gas at a distance of 0.7 light-years from the star that exploded. Calculate the speed of that debris. Give the answer in both light-years per year and km/hr or miles/hr, whichever is more intuitive for you. How does the speed you find compare with the speed of light?
1. The speed (in light-years per year) of the debris is 4.67×10⁻² light-years / year
2. The speed (in Km/h) of the debris is 50436000 Km/h
3. The speed of light is 21.4 times the speed of the debris
What is speed?Speed is the distance travelled per unit. Mathematically, it can be expressed as:
Speed = distance / time
How to determine the speed in light-years per yearDistance = 0.7 light-yearsTime = 15 years Speed = ?Speed = distance / time
Speed = 0.7 / 15
Speed = 4.67×10⁻² light-years / year
2. How to determine the speed in Km/hSpeed (in light-years per year) = 4.67×10⁻² light-years / yearSpeed (in Km/h) = ?1 light-years / year = 1.08×10⁹ Km/h
Therefore,
4.67×10⁻² light-years / year = 4.67×10⁻² × 1.08×10⁹
4.67×10⁻² light-years / year = 50436000 Km/h
3. Comparison of the speed of the debris with the speed of lightSpeed of light = 3×10⁸ m/sSpeed of debris = 50436000 Km/h = 50436000 / 3.6 = 14010000 m/sComparison =?Comparison = Speed of light / speed of debris
Speed of light / speed of debris = 3×10⁸ / 14010000
Speed of light / speed of debris = 21.4
Cross multiply
Speed of light = 21.4 × Speed of debris
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It takes 23 hours 56 minutes and 4 seconds for the earth to make one revolution (mean sidereal day). What is the angular speed of the earth?
Answer: 7.29×10^-5 rad/s
Assume the earth is spherical. Relative to someone on the rotation axis, what is the linear speed of an object on the surface if the radius vector from the center of the earth to the object makes an angle of 49.0° with the axis of rotation. The radius of the earth is 6.37×103 km.
Answer: 3.51×10^2 m/s
What is the acceleration of the object on the surface of the earth in the previous problem?
This is the question I need the answer for.
The angular speed of the earth, the linear speed of an object on the surface, and the acceleration of the object will be 7.288 × 10⁻⁵ m/sec,446.36 m/sec, and 31.27 m/s² respectively.
What is acceleration?The rate of velocity change concerning time is known as acceleration.
Unit conversion;
1 hour = 3600 sec
Given data;
Velocity, v= m/s
Time elapsed, t = 23 hours 56 minutes and 4 seconds
The radius of the earth is, R= 6.37×103 km.
The total taken in the second is;
T=23 hr × 3600 sec + 56 min × 60 sec + 4 sec
T= 86164 sec
The angular speed of the earth;
[tex]\rm \omega_e = \frac{2 \pi}{T} \\\\ \omega_e =\frac{ 2 \times 3.14 }{86164 \ sec} \\\\ \omega_e =7.28 8 \times 10^{-5 } \ rad /sec[/tex]
The linear speed of an object on the surface of the radius vector from the center of the earth is;
[tex]\rm v = r \times \omega \\\\ v= 6123 \ km \times 7.29 \times 10 ^{-5} \\\\ v = 446.36 \ m/sec[/tex]
The acceleration of the object on the surface of the earth is;
[tex]\rm a = \frac{v^2}{r} \\\\ a=\frac{4446.32^2}{6.37 \times 10^3} \\\\ a= 31.27 \ m/s^2[/tex]
Hence,the angular speed of the earth, the linear speed of an object on the surface, and the acceleration of the object will be 7.288 × 10⁻⁵ m/sec,446.36 m/sec, and 31.27 m/s²
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Which would most likely cause a decrease in the rate of energy production in a fusion nuclear reactor?
In a fusion nuclear reactor, a drop in temperature would most likely result in a decline in the rate of energy output.
What is nuclear power?The utilization of nuclear reactions to generate energy is known as a nuclear power. Nuclear fission, nuclear decay, and nuclear fusion processes are all sources of nuclear energy.
Nuclear power facilities currently produce the great bulk of the electricity generated by nuclear fission of uranium and plutonium.
The environment must be hot enough for the deuterium and tritium ions' kinetic energies to be sufficient to break through the Coulomb barrier and fuse together.
Hence a decrease in temperature would most likely cause a decrease in the rate of energy production in a fusion nuclear reactor.
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Answer:
A: a drop in the temperature in the reactor
Explanation:
right on edg
A uniform rod with a mass of m = 1.94 kg and a length of l = 2.10 m is attached to a horizontal surface with a hi=nge. The rod can rotate around the hi=nge without friction. (See figure.)
Initially the rod is held at rest at an angle of θ = 70.4° with respect to the horizontal surface. Then the rod is released.
What is the angular speed of the rod, when it lands on the horizontal surface?
1.61 rad/s is incorrect.
What is the angular acceleration of the rod, just before it touches the horizontal surface?
1.06 rad/s^2 is incorrect.
The angular acceleration of the rod, just before it touches the horizontal surface is mathematically given as
w = 3.6 rad/s
What is the angular acceleration of the rod, just before it touches the horizontal surface?
Generally, the equation for Torque is mathematically given as
[tex]T = F r[/tex]
Therefore
[tex]T = (mg Cos\theta) (L/2)\\\\T = I \alpha[/tex]
[tex]\alpha = (1.5) (9.8 Cos65.2) (2.10)[/tex]
[tex]\alpha = 12.95 rad/s2[/tex]
Generally, the equation for conservation of energy is mathematically given as
P.E=K.E
Therefore
[tex]mg (L/2) sin65.2 = (0.5) I w2[/tex]
[tex](0.5) (9.8) Sin65.2 = (0.5) (2.1/3) w2[/tex]
w = 3.6 rad/s
In conclusion, the angular acceleration
w = 3.6 rad/s
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Three objects are brought close to each other, two at a time. When objects A and B are brought together, they attract. When objects B and C are brought together, they repel. From this, we conclude that (a) objects A and C possess charges of the same sign. (b) objects A and C possess charges of opposite sign. (c) all three of the objects possess charges of the same sign. (d) one of the objects is neutral. (e) we need to perform additional experiments to determine information about the charges on the objects.
B. Objects A and C possess charges of opposite sign.
Attraction of two objects
The attraction of two objects occurs when the two objects have opposite charges.
Repulsion of two objectsThe repulsion of two objects occurs when the two objects have same charges.
Thus, we can conclude that, objects A and C possess charges of opposite sign.
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A dolphin accelerates at -1.77 m/s²
for 3.33 s, and ends with a velocity
of -8.77 m/s. What is the displacement
of the dolphin in that time?
Which of the following changes would double the force between two charged particles? O A. Decreasing the distance between the particles by a factor of 2 B. Increasing the distance between the particles by a factor of 2 O C. Doubling the amount of charge on one of the particles O D. Doubling the amount of charge on each particle
Let's check
[tex]\\ \rm\dashrightarrow F=\dfrac{k}{q_1q_2}{r^2}[/tex]
[tex]\\ \rm\dashrightarrow F\propto Q[/tex]
[tex]\\ \rm\dashrightarrow F\propto \dfrac{1}{r^2}[/tex]
So
Option A and C can be used
Answer:
Decreasing the distance between the particles by a factor of 2
Explanation:
To double the force between the two charge particles, the distance between the particles should be reduced by a factor of two.
According to coulombs law "force is directly proportional to the potential between the two charges and inversely proportional to the square of their distances".
[tex]\sf{F = \dfrac{kq_{1} q_{2} }{r^{2} }}[/tex]
F is the electric force
k is the coulomb constant
q is the charge
r is the distance or separation
As the separation between the charges is reduced the force increases and vice versa.
A firefighter is using a hose and the flow rate of the water leaving the hose is 0.032 m3/s. At the end of the hose, the nozzle has a radius of 0.013 m. What is the speed of the water leaving the hose ?
A firefighter is using a hose and the flow rate of the water leaving the hose. the speed of the water leaving the hose is mathematically given as
u = 9.947 m/s
What is the speed of the water leaving the hose?
Generally, the equation for Volume flow rate is mathematically given as
(V)= (A) * (u)
Where
A= area
v=velocity
Therefore
V = A*u
V=0.032 m3/s
A=πR2
A= π*0.013
u=V/A
u=0.032/π*0.0322
u = 60.27m/s
In conclusion, the speed of the water leaving the hose
u = 60.27m/s
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The radius of curvature of a highway exit is r = 93.5 m. The surface of the exit road is horizontal, not banked. (See figure.)
If the coefficient of static friction between the tires of the car and the surface of the road is μ[tex]_{s}[/tex] = 0.402, then what is the maximum speed at which the car can safely exit the highway without sliding?
Static friction keeps the car from skidding off the road and points toward the center of the curve. By Newton's second law, the car experiences
• net vertical force
F [normal] - F [weight] = 0
• net horizontal force
F [friction] = ma = mv²/r
where v is the tangential speed of the car.
It follows that
F [normal] = F [weight] = mg
and when static friction is maximized at the car's maximum speed,
F [friction] = µ F[normal] = 0.402 mg
Solve for v :
0.402 mg = mv²/r ⇒ v = √(0.402 g (93.5 m)) ≈ 19.2 m/s
Lesson 2 History of Physical Science
Write an expository essay explaining how science builds on itself. Use at least two specific examples from this lesson.
The prompt above requires an explanatory essay. The focus of an explanatory essay is to give clarity to a concept. See the sample below.
What is the explanatory essay as per the prompt above?From history, it is clear that Science builds on itself. For avoidance of doubt, this means that improvements in science are incremental with each improvement feeding off the last.
From the lesson we have examples of this phenomena such as that of Galileo. Recall that the text indicates that He improved the telescope; and while carrying out the same research that his predecessor Copernicus had done in relation to the stars and planets, Galileo used math to justify his position.
It is also on record that Johannes Kepler who lived in the same period as Galileo took interest in Astronomy. In this case, science built on itself because, Kepler's work served as the foundation for Sir Isaac Newton's work. This is also another example of how science builds on itself.
From the above textual evidence, it is given beyond reasonable double that science indeed builds on itself with the discovery of each generation serving as the foundation for the advancement of the next.
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What Is sonar method ?
[tex]\huge\underline{\red{A}\green{n}\blue{s}\purple{w}\pink{e}\orange{r} →}[/tex]
Sonar is an abbreviation of Sound Navigation and Ranging
It is used to detect objects submerged under water such as sea rocks and submarines, and for measuring their depth.Ultrasonic waves from a transmitter are sent into the water. When intercepted by some rock or submarine, they are reflected back and detected by a detector. Measuring the time–interval between the transmission and reception of the Ultrasonic waves and knowing their speed in sea water, the depth of the rock or the submarine can be calculated. The principle of Sonar is also used in detecting flaws in materials and in mechanical testing of materials without damaging them.Explanation:
Hope it helps you!!If argon could exist as a solid, what would best represent the speed through solid argon? 10 m/s 250 m/s 500 m/s 3200 m/s
3200 m/s best represents the speed through solid argon.Option D is correct.
What is a sound wave?A sound wave is produced when a medium begins to vibrate. When an entity vibrates, a pressure wave is formed, which causes sound.
Through argon, sound travels at a speed of 319 m/sec.
Any chemical element in the fluid phase has one temperature-dependent value for the speed of sound.
Varying sound wave types may travel through solids at different speeds. These wave types include longitudinal transverse, and extensional.
If argon could exist as solid, then 3200 m/s is the best speed.
Hence, option D is correct.
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A knife is dropped from the top of a 10.0 m high mast on a ship moving at 1.74 m/s due south.
(a) Calculate the velocity of the knife relative to the ship when it hits the deck of the ship.
m/s (down)
(b) Calculate the velocity of the knife relative to a stationary observer on shore.
m/s
° (below the horizontal to the south)
(c) Discuss how the answers give a consistent result for the position at which the knife hits the deck.
a)The velocity of the knife relative to the ship when it hits the deck of the ship will be 14.00 m/sec.
b)The velocity of the knife relative to a stationary observer on shore is 14.10 m/sec
c)The position at which the knife hits the deck is 82.8°.
What is velocity?The change of displacement with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is;
The velocity of the knife relative to the ship when it hits the deck of the ship is;
v=√2gh
v=√2×9.81 ×10.0 m
v=14.00 m/sec
b)
The velocity of the knife relative to a stationary observer on shore is found as;
v₁₂ = √[(14.00)²+(1.74)²]
v₁₂=14.10 m/sec
c)
The position at which the knife hits the deck is;
[tex]\rm \theta = tan^{-1}\frac{14.00}{1.75} \\\\ \theta = 82.8 ^ 0[/tex]
Hence, the velocity of the knife relative to the ship, the velocity of the knife relative to a stationary observer on shore, and the position at which the knife hits the deck will be 14.00 m/sec,14.10 m/sec, and 82.8° respectively.
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PLSSSS Jenny runs every morning for exercise she takes 25 minutes to run 1,000 m and walk 1,600 m or 20 minutes to run 2,000 m and walk 800 m if her running speed and walking speed do not change what is her running speed and how long does she take to walk 800 m
The running speed of Jenny willl be 2.5 times the velocity she walks.
What is velocity?The change of distance concerning to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
Condition 1;
She takes 25 minutes to run 1,000 m and walk 1,600 m The total time to run as well as the walk is 25 minutes.
[tex]\rm t = t_r +t_w \\\\ 25 = t_r +t_w[/tex]
As we know that;
Distance = speed × time
[tex]\rm x_r = v_r \times t_r \\\\ t_r = \frac{x_r}{v_r}[/tex]
For walking;
[tex]\rm t_w = \frac{x_w}{v_w}[/tex]
[tex]\rm t = t_r +t_w \\\\ 25 = \frac{x_r}{v_r} +\frac{x_w}{v_w} \\\\ 25 = \frac{1000 \ m}{v_r} +\frac{1600 \ }{v_w} \\\\ 40 v_w +64 v_r = v_rv_w[/tex]
Same as for condition 2;
[tex]\rm 100 v_w + 40 v_r = v_rv_w[/tex]
[tex]\rm 40 v_w +64 v_r =\rm 100 v_w + 40 v_r \\\\ v_r = 2.5 v_w[/tex]
Condition 3;
[tex]\rm d_w = 800 \ m[/tex]
[tex]\rm d_w = v_w \times t_w[/tex]
[tex]\rm t_w = \frac{d_w}{v_w } \\\\ \rm t_w = \frac{800}{v_w } \\\\[/tex]
Hence their running speed of Jenny willl be 2.5 times the velocity at she walks.
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What function is the opposite of observant in the 16 personality test
[tex]\Huge\boxed{\textsf{Intuitive}}[/tex]
The intuitive and observant traits are meant to describe what people do with the information they gather from the world around them.
Observant people are more interested in observable facts and straightforward outcomes, while intuitive people imagine the future potential of what they see.
For example, intuitive people are more likely to discuss views and theories of what the world could be like in the future.
Answer:
intuition
Explanation:
Thermal_____________ is applied when objects are at equal temperature.
O conduction
O convection
O equilibrium
Answer:
Equilibrium
Explanation:
Temperature is a measure of the average kinetic energy of the atoms or molecules in the system.
The zeroth law of thermodynamics states that there is no heat transferred between two objects in thermal equilibrium
This means that they are the same temperature.
I need help with my homework
The center of mass of the boat will change to 0 meters.
What is the center of mass?A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.
Let's think of the rowboat and man as a system. From one end of the boat to the other, the individual is walking.
To preserve momentum, the rowboat advances in the opposite direction as the person begins to move from, say, the left end to the right end.
As a result, the system's center of mass doesn't move because there isn't any external force acting on it.
Hence, the center of mass of the boat will change to 0 meters.
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The figure below shows a dipole. If the positive particle has a charge of 37.3 mC and the particles are 3.08 mm apart, what is the electric field at point A located 2.00 mm above the dipole's midpoint? (Express your answer in vector form.)
The electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.
Electric field of the positive particle
The electric field is calculated as follows;
E = kq/r²
where;
r is the distance between the chargesk is Coulomb's constantq is magnitude of the chargemidpoint of 3.08 m, x = 1.54 mm
r(1.54 mm, 2.00 mm)
|r| = √(1.54² + 2²)
|r| = 2.52 mm
E = (9 x 10⁹ x 37.3 x 10⁻³)/(2.52 x 10⁻³)²
E = 5.287 X 10¹³ N/C
Thus, the electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.
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