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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Two football players collide head-on in midair while chasing a pass. The first player has a 101.0 kg mass and an initial velocity of 4.10 m/s, while the second player has a 117 kg mass and initial velocity of -6 m/s. What is their velocity (in m/s) just after impact if they cling together? (Indicate the direction with the sign of your answer.)
Assuming players stick together immediately after the collision, their speed will be 1.3206 m/sec, and - ve indicates their direction.
What is the law of conservation of momentum?According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
(m₁) is the mass of 1st player= 101.0 kg
(u₁) is the initial velocity of 1st player (u₁) = 4.10 m/s
(m₂) is the mass of 2nd player = 117 kg
(u₂) is the initial velocity of 2nd player= -6 m/s
(v) is the velocity after collision =.?
According to the law of conservation of momentum;
Momentum before collision =Momentum after collision
m₁u₁+m₂u₂=(m₁+m₂)V
101.0 kg × 4.10 m/sec + 117 kg × (-6 m/sec) = (101.0 kg + 117 kg) × V
V= - 1.3206 m/sec
Hence, the velocity of players just after impact, if they cling together, will be - 1.3206 m/sec, and - ve shows the direction.
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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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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
Discuss the causes, characteristics, and major features of black holes. Explain why the concept of a black hole does or doesn't seem reasonable to you and provide some rationale for your views?
The concept of a black hole seem reasonable to me because of its valid and authentic evidences.
What are the causes, characteristics, and major features of black holes?A black hole is a region in space where light is unable to escape due to strong gravitation force. The strong gravity is formed because matter has been pressed into a tiny space. This compression occurs at the end of a star's life. Some black holes are formed when a stars die.
So we can conclude that the concept of a black hole seem reasonable to me because of its valid and authentic evidences.
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DNA is unique to each __________.
Answer:
ORGANISM
Explanation:
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?
764.8 N is incorrect.
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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M, a solid cylinder (M=2.15 kg, R=0.111 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.730 kg mass, i.e., F = 7.161 N. If instead of the force F an actual mass m = 0.730 kg is hung from the string, find the angular acceleration of the cylinder.
The angular acceleration of the cylinder is mathematically given as
[tex]\alpha = 60.2175 rad/s^2[/tex]
What is the angular acceleration of the cylinder.?Question parameters
M=2.15kgR = R=0mW= 7.161 NGenerally, the equation for Torque generated in the cylinder is mathematically given as
[tex]\sigma[/tex] = w×r
for circular motion torque is
[tex]\sigma = × \alpha[/tex]
t[tex]=0.5 × m ×r^2 × \alpha\\\\ =0.5 × 2.15 ×0.111 ^2 × \alpha[/tex]
t = 0.0132 ×[tex]\alpha[/tex]
Therefore
0.0132 × [tex]\alpha[/tex] = w × r
[tex]\\\\\alpha =\frac{ 7.161 0.111}{(0.0132)}[/tex]
[tex]\alpha = 60.2175 rad/s^2[/tex]
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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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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 net external force is exerted on a 1350-kg artillery shell fired from a battleship if the shell is accelerated at 2.00 ✕ 104 m/s2? (Enter the magnitude.)
N
What is the magnitude of the force exerted on the ship by the artillery shell?
N
The magnitude of the force exerted on the ship by the artillery shell will be 27× 10⁶ N.
What is force?Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.
Force is defined as the product of mass and acceleration. Its unit is Newton.
Given data;
Force,F = ?
Mass,m = 1350-kg
Acceleration,a = 2.00 ✕ 10⁴ m/s²
The magnitude of the force exerted on the ship by the artillery shell is found as;
F=ma
F=1350-kg × 2.00 ✕ 10⁴ m/s²
F= 10,000 N
The magnitude of the force exerted on the ship by the artillery shell will be 27× 10⁶ N.
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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 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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a= 3t^3 + 2t^2 find
1) velocity at t= 2s.
2) displacement at 4s
please answer this question ASAP
Use the fundamental theorem of calculus.
[tex]f(t) = f(0) + \displaystyle \int_0^t f'(u) \, du[/tex]
In order to find velocity and position exactly, you need to know the initial velocity and position.
1) Integrate the acceleration function to get the velocity function. By the FTC,
[tex]v(t) = v(0) + \displaystyle \int_0^t a(u) \, du[/tex]
[tex]v(t) = v(0) + \displaystyle \int_0^t (3u^3+2u^2) \, du[/tex]
[tex]v(t) = \dfrac34 t^4 + \dfrac23 t^3 + v(0)[/tex]
At [tex]t=2\,\rm s[/tex], the velocity is
[tex]v(2) = \dfrac34 2^4 + \dfrac23 2^3 + v(0) = \boxed{\dfrac{52}3 + v(0)}[/tex]
2) Integrate the velocity function to the get the position function. By the FTC again,
[tex]x(t) = x(0) + \displaystyle \int_0^t v(u) \, du[/tex]
[tex]x(t) = x(0) + \displaystyle \int_0^t \left(\frac34 u^4 + \frac23 u^3 + v(0)\right) \, du[/tex]
[tex]x(t) = \displaystyle \frac3{20} t^5 + \frac16 t^4 + v(0) t + x(0)[/tex]
At [tex]t=4\,\rm s[/tex], the object's position is
[tex]x(4) = \dfrac3{20}4^5 + \dfrac16 4^4 + 4v(0) + x(0) = \dfrac{2944}{15} + 4v(0) + x(0)[/tex]
Fortunately, you don't need the initial position to find the displacement, since [tex]x(0)[/tex] cancels out in the end.
[tex]\Delta x = x(4) - x(0) = \boxed{\dfrac{2944}{15} + 4v(0)}[/tex]
A stone is dropped from rest into a well. The sound of the splash is heard exactly 2.70 s later. Find the depth of the well if the air temperature is 11.0°C.
The depth of the well at the given air temperature is 456.3 m.
Depth of the wellThe depth of the well is calculated from the principle of echo.
v = 2d/t
2d = vt
d = vt/2
where;
v is speed of sound in air at 11.0°C = 338 m/sd = (338 x 2.7)/2
d = 456.3 m
Thus, the depth of the well at the given air temperature is 456.3 m.
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A Skier starting from rest rolls down a 35.0 m ramp. When he arrives at the bottom of the ramp his
speed is 10.43 m/s. (a) Determine the magnitude of his acceleration, assumed to be constant, (b) If the
ramp is inclined at 60.0 degrees with respect to the ground, what is the component of his acceleration
that is parallel to the ground. (c) Provide me with a free body diagram, one for the 35 m ramp situation.
Also, provide a nice sketch of the skier.
The acceleration is 1.55 m/s^2 while is component that is parallel to the ground is 0.775 m/s^2.
What is acceleration?The term acceleration is the rate of change of speed. Given that the distance covered is 35.0 m and the final velocity is 10.43 m/s it the follows that;
v^2 = u^2 + 2as
u = 0 because he started from rest
v^2 = 2as
a = v^2/2s
a = (10.43 m/s)^2/2 * 35.0 m
a = 1.55 m/s^2
The component of this acceleration which is parallel to the ground is;
ax = a sin θ
ax = 1.55 m/s^2 cos 60
ax = 0.775 m/s^2
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A cannonball is shot horizontally off a
10.5 m high castle wall at 47.4 m/s. How
far from the base of the wall does the
cannonball land?
Answer:it's 132.2m
Explanation:
Hello !
Because first recall parabolic peak height equation and
given that v=47.4m/s , h=10.5m
v²sin²(θ⁰)=2ℎ₀(47.4)²²(θ⁰)=2(9.81)(10.5) θ=17.63secondly recall parabolic total distance equation
d₀=²(2θ⁰)/gd₀=(47.4)²(2(17.63)⁰)/9.81d₀=132.2mThe distance from the base of the wall that the cannonball land will be 138.7 m.
What is the peak height equation?Consider that a body of mass 'm' is projected with a velocity of projection 'v' and angle of projection 'θ'. Assume that 'h' be the maximum height attained by the projectile.
2gh = v²sin²θ
This expression can be used to calculate the maximum height of the projectile.
Given, the height of the castle wall, h = 10.5 m
The velocity of the cannonball, v = 47.4m/s
The peak height equation can be used to find the angle θ:
v²sin²θ = 2gh
(47.4)² sin²θ = 2(9.81)(10.5)
Sinθ = 0.3028
θ = 17.63°
Therefore, the distance (d) from the base of the wall that the cannonball land:
d = (47.4)² × 2 sin(17.63) / 9.81
d = 138.7m
Therefore the distance from the base of the wall that the cannonball land is 138.7m.
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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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At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40 degrees above horizontal. She leaves the cannon at 2m above ground and lands in a net 1m above the ground. At what height does a human cannonball reach?
The maximum height reached by the human cannonball is 4.74 m.
Height reached by the human cannonballThe height reached is calculated as follows;
H = u²sin²θ/2g
where;
u is the initial velocityθ is the angle of projectionH = (15² x [sin(40)]² )/(2 x 9.8)
H = 4.74 m
Thus, the maximum height reached by the human cannonball is 4.74 m.
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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
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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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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In 450 BCE, Greek philosopher Democritus first thought of the existence of tiny particles that compose everything around us. He named them 'atomos', meaning _________.
A. None of these
B. Indivisible
C. Invisible
D. Particle
B. The meaning of 'atomos' according to Democritus in 450 BCE is indivisible.
The building block of every matter
A Greek philosopher known as Democritus first thought of the existence of tiny particles that compose everything around us.
Democritus of Abdera, named the building blocks of matter atomos, meaning literally “indivisible.
Thus, the meaning of 'atomos' according to Democritus in 450 BCE is indivisible.
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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
Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.
Calculate the magnitude of the tension in the string marked A. (You'll need to get the various positions from the graph. The ends of the strings are exactly on one of the tic marks.)
The magnitude of the tension in the string marked A and B is mathematically given as
A = 52.5 NB = 39.5 NWhat is the magnitude of the tension in the string marked A?Generally, the equation for is mathematically given as
The angle at A is...
[tex]tan\theta = 3/8[/tex]
When below the negative x.
B= tan\phi
[tex]tan\phi = 5/4[/tex]
When below the negative x
C=
[tex]tan \rho = 1/6[/tex]
Hence, considering the Horizontal components
74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)
A = 78.9 - 0.668B
Hence, considering the Vertical components
74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)
40.07 = 1.015B
B = 39.5 N
In conclusion, Sub the value of B is the equation of A
A = 78.9 - 0.668B
Sub
A = 78.9 - 0.668( 39.5 N)
A = 52.5 N
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Many immigrants lived in
A school bus is moving 26.8 m/s on
flat ground when it begins to
accelerate at 4.73 m/s². How much
time does it take to travel 95.1 m?
(Unit=s)
Watch your minus signs!
time (s)
Enter
The time taken for the bus to travel 95.1 m is 2.8 s
How to determine the final velocityInitial velocity (u) = 26.8 m/sAcceleration (a) = 4.73 m/s²distance (s) = 95.1 mFinal velocity (v) = ?v² = u² + 2as
v² = 26.8² + (2 × 4.73 × 95.1)
v² = 1617.886
Take the square root of both sides
v² = √1617.886
v = 40.2 m/s
How to determine the timeInitial velocity (u) = 26.8 m/sAcceleration (a) = 4.73 m/s²Final velocity (v) = 40.2 m/sTime (t) = ?
v = u + at
40.2 = 26.8 + (4.73 × t)
Collect like terms
4.73 × t = 40.2 - 26.8
4.73 × t = 13.4
Divide both side by 4.73
t = 13.4 / 4.73
t = 2.8 s
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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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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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A
В
C
D
At which point will an image be formed?
ОА
Ов
Ос
OD
Answer:
oa
Explanation:
it may be oa is the right answer for this question
but I don't know properly
In which part of the ear is the sound wave converted into an electrical impulse?