Answer:
Explanation:
a ) distance s = 12 m .
constant speed = 1.5 m/s
time taken to coast 12 m
= distance / speed
= 12 / 1.5 = 8 s
b ) initial velocity u = 2.5 m /s
final velocity v = 1.5 m /s
displacement s = 14 m
acceleration a = ?
v² = u² + 2 as
1.5² = 2.5² + 2 a x 14
a = - .1428 m /s²
= - 14.28 cm / s²
c )
v = ?
u = 2.5 m /s
s = 6 m
a = - .1428 m /s²
v² = u² - 2 as
= 2.5² - 2 x 6 x .1428
= 6.25 - 1.71
= 4.54
v = 2.13 m /s .
Suppose that the estimate of the distance that the ball moves while a player is throwing it is 2.5 feet -- that is, the distance from where the ball is when the player starts their throw until it leaves their hand. Calculate the magnitude of the average acceleration in m/s2m/s2 that the ball has while it is being thrown, as it moves from rest to the point where it leaves the player's hand.
This question is incomplete. the complete question is;
a) Suppose that the estimate of the maximum height that a player can throw a baseball is 25 ft. For how long after the ball leaves the player's hand does it return to the players hand?
b) Suppose that the estimate of the distance that the ball moves while a player is throwing it is 2.5 feet -- that is, the distance from where the ball is when the player starts their throw until it leaves their hand. Calculate the magnitude of the average acceleration in m/s2m/s2 that the ball has while it is being thrown, as it moves from rest to the point where it leaves the player's hand.
Answer:
a) Total time of Flight is 2t is 2.49 seconds
b) The magnitude of the average acceleration is 98.03 m/s²
Explanation:
Given the data in the question;
a)
Maximum height [tex]h_{max}[/tex] = 25 ft = (25 /3.281) = 7.62 m
For how long after the ball leaves the player's hand does it return to the players hand;
[tex]h_{max}[/tex] = [tex]\frac{1}{2}[/tex]gt²
we know that g = 9.81 m/s²
so we substitute
7.62 m = [tex]\frac{1}{2}[/tex] × 9.81 m/s² × t²
7.62 m = 4.905m/s² × t²
t² = 7.62 m / 4.905
t² = 1.5535
t = √1.5535
t = 1.246 s
So total time of Flight is 2t = 2 × 1.246 s = 2.49 seconds
b)
given that
distance = 2.5 ft = ( 2.5 / 3.281) = 0.762 m
V = ?
from the First Equation of Motion
v = u + at
0 = u - 9.81 × 1.246 s
u = 12.2232 m/s
so from the Third Equation of Motion : v² = u² + 2as
(12.2232 m/s)² = 0² + 2 × a × 0.762 m
149.4066 = 1.524a
a = 149.4066 / 1.524
a = 98.03 m/s²
Therefore, the magnitude of the average acceleration is 98.03 m/s²
what is the acceleration of an object that has a mass of 10kg and is pushed with a force of 50 N. just answer this question with the number no unit
Answer:
a = 5 [m/s²]
Explanation:
We know from Newton's second law that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = total force = 50 [N]
m = mass = 10 [kg]
a = acceleration [m/s²]
Now replacing:
[tex]a = \frac{F}{m} \\a=50/10\\a = 5 [m/s^{2} ][/tex]
A tank, in the shape of a cube with each side measuring 2 meters, is half-full of water. Above the water is air that is pressurized to 5 kPa. Calculate the resultant force on one side of the tank due to the air and water (ignore atmospheric pressure) and determine the height of this force above the bottom of the tank.
Answer:
Explanation:
The tank is half full so height of water level is at 1 m . Centre of gravity of water will be at height of .5 m . Pressure will act at this point .
Pressure = h d g where h is height of centre of mass of water column , d is density of water and g is acceleration due to gravity .
Pressure of water column = .5 x 10³ x 9.8 = 4.9 k Pa .
Air is pressurized to 5 kPa so
resultant pressure on one side of the tank due to the air and water
= 4.9 + 5 kPa = 9.9 kPa .
Total force on one face = pressure x area of one face under water
= 9.9 x 10³ x .5 x 2²
= 19.8 kN .
On a winter day a child of mass 20.0 kg slides on a horizontal sidewalk covered in ice. Initially she is moving at 3.00 m>s, but due to friction she comes to a halt in 2.25 m. What is the magnitude of the constant friction force that acts on her as she slides
Answer:
40 N
Explanation:
According to the scenario, computation of given data are as follows:
Mass (m) = 20 kg
Initially moving (v) = 3
Actual distance (d) = 2.25 m
So, we can calculate friction (f) by using following formula,
f × d = [tex]\frac{1}{2} mv^{2}[/tex]
By putting the value, we get
f × 2.25 = [tex]\frac{1}{2}[/tex] × 20 × [tex]3^{2}[/tex]
f × 2.25 = 10 × 9
f = 90 ÷ 2.25
= 40 N.
__________ forces are not equal, and they always cause the motion of an object to change the speed and/or direction of an object.
Answer:
Unbalanced Forces.
Explanation:
Answer:
Unbalanced
Explanation:
Unbalanced forces are not equal and they always cause the motion of an object to change the speed and or direction of the object.
Balanced forces are equal and will not cause the motion of an object to change.
From Newton's law of inertia "an object will continue in its state of rest or of uniform motion unless acted upon by an external force".
The net force for a body with unbalanced forces is greater than zero.
For a body with balanced forces, the net force is zero
On a particular day the ocean swell has a period of 10 s, a wave velocity of 5.0 m/s, and a wave amplitude of 0.520 m. An anchored boat is bobbing up and down as the wave passes by. What is the peak vertical velocity of the boat
Answer:
the peak vertical velocity of the boat is 0.3267 m/s
Explanation:
Given that;
Time period T = 10 sec
wave velocity = 5.0 m/s
wave amplitude A = 0.520 m
peak vertical velocity of in the wave can be expressed as;
Vp = Aw
where w is angular frequency, ( w = 2π/T)
so
Vp = A × 2π/T
so we substitute
Vp = 0.520 × 2π/10
Vp = 3.2673 / 10
Vp = 0.3267 m/s
Therefore, the peak vertical velocity of the boat is 0.3267 m/s
One disadvantage to experimental research is that experimental conditions do not always reflect reality.
Please select the best answer from the choices provided
T
F
Answer:
It's true I took the test on Edge.
Explanation:
Answer:
True
Explanation:
Got it right on edg
A student releases a small cart at the top of an incline with height H above the floor. The cart experiences very little friction. The student is attempting to cause the cart to go around a vertical loop of radius R without the cart losing contact with the track at the top. The student suggests that the heigt H should equal 2R so that the release height and maximum height of th eloop are the same. However, the student finds that it requires noticably higher hieght than 2R for the cart to go around the loop. Explain why H must be noticably greater than 2R to complete the loop. (Hint: In order for the cart to go around the loop it must have a nonzero velocity at the top of the loop.) answer
Answer:
Explanation:
In the whole process , potential energy of the cart is converted into kinetic energy . At the top of the vertical loop , the whole of potential energy is regained and kinetic energy becomes zero if we release the cart from a height of 2R because difference of height between lowest and highest point of motion is 2R . In that case kinetic energy at top = 0 , velocity v = 0
At the top , weight mg is acting which is providing centripetal force . So cart must have some velocity at the top . If it be v
mv²/R = mg
v = √ gR .
For that purpose , the cart must be released from a height greater than 2R .
The extra height beyond 2R will make the velocity at the top non-zero.
I don’t even understand anyone help please.
Answer:
a) A:170572.5 J
C: 55794.9J
b) 170572.5 J
c) 41.4413265306m
d) 2.7455874717m/s
Explanation:
a) Kinetic energy = 0.5*m*v²
KE at A = 0.5*420*28.5² = 170572.5 J
KE at C = 0.5*420*16.3² = 55794.9 J
b) Mechanical energy is the total kinetic energy plus potential energy at a point on the system. There is no potential energy at A.
ANSWER: 170572.5 J
c) v²=u²+2as
28.5²=2(9.8)s
812.25/19.6=s
s=41.4413265306m
d) h=height from part c, r=radius of loop
v²=u²+2as
v²=gr or a=v²/r
Ei=Ef
mgh=0.5mv²+mg(2r)
gh=0.5v²+2gr
h=0.5r+2r
h=5/2r
r=2/5h=(2/5)(41.4413265306)=16.5765306122
F=ma
mg=m(v²/r)
g=v²/r
v²=(9.8)(16.5765306122)
v=√162.45
=12.7455874717m/s
What is physical education mean
Answer:
Ps eso me salio
Physical education, sports education or sports education are terms that refer to the teaching and learning of physical exercises whose main objective is education and health. This has been the decisive reason for the introduction of physical exercises in elementary school in the 19th century.
Explanation:
The impulse given to a body of mass 1.5 kg, is 6.0 kg
• m•s-?
If the body was initially at rest, what
will its resulting kinetic energy be? Give your answer in J without units.
Show work
Answer:
12J
Explanation:
Given parameters:
Mass = 1.5kg
Impulse = 6kgm/s
let us start first by find the velocity with which this body moves;
Impulse = mass x velocity
Velocity = Impulse / mass = 6/ 1.5 = 4m/s
Initial velocity = 0m/s
Unknown:
Resulting kinetic energy = ?
Solution:
To solve this problem use the formula below:
K.E = [tex]\frac{1}{2}[/tex] m (v - u)²
m is the mass
v is the final velocity
u is the initial velocity
So;
K.E = [tex]\frac{1}{2}[/tex] x 1.5 x (4 - 0)²
K.E = 1.5 x 8 = 12J
A 1200 kg truck accelerates from 25 m/s to 53 m/s over a distance of 350 m.
What is the average net force on the truck?
Answer:
the average net force on the truck is 3744 N.
Explanation:
Given;
mass of the truck, m = 1200 kg
initial velocity of the truck, u = 25 m/s
final velocity of the truck, v = 53 m/s
distance traveled by the truck, d = 350 m
The acceleration of the truck is calculated as;
v² = u² + 2ad
53² = 25² + (2 x 350)a
700a = 53² - 25²
700a = 2184
a = 2184 / 700
a = 3.12 m/s²
The average net force on the truck is calculated as;
F = ma
F = 1200 x 3.12
F = 3744 N
Therefore, the average net force on the truck is 3744 N.
An airplane is heading due south at a speed of 560 km/h . If a wind begins blowing from the southwest at a speed of 80.0 km/h (average). Calculate magnitude of the plane's velocity, relative to the ground.
Answer:
the magnitude of Vpg = 493.711 km/h
Explanation:
given data
speed Vpg = 560 km/h
speed Vwg = 80 km/h
solution
we get here magnitude of the plane velocity w.r.t. ground is
we know that the Vpg = Vpw + Vwg .....................1
writing the component of the velocity that is
Vpw = (0 km/h î - 560 km/h j )
Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)
adding these
Vpg = (0+80 cos 45 km/h ) î + ( -560 + 80 sin 45 km/h j)i
Vpg = (42.025 ) î (-491.92 km/h)j
now we take magnitude
the magnitude of Vpg = [tex]\sqrt{(42.025^2+(-491.92)^2)} km/h[/tex]
the magnitude of Vpg = 493.711 km/h
A salmon jumps up a waterfall 2.4 m high. With what minimum speed did the salmon leave the water below to reach the top?
Answer:
6.86 m/s
Explanation:
The minimal velocity needed is when we have only vertical motion, then i will think in the problem only in one axis.
I suppose that the only force, in this case, is the gravitational force acting on the fish.
Then the gravitational equation of the fish will be:
a(t) = -9.8m/s^2
For the velocity equation we need to integrate over time to get:
v(t) = (-9.8m/s^2)*t + v0
Where v0 is the initial velocity of the fish and is what we want to find.
For the position equation we need to integrate over time again to get:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + v0*t + p0
p0 is the initial position of the fish, and because he starts one the water, the initial position is p0 = 0 m
Then the equation is:
p(t) = (1/2)*(-9.8 m /s^2)*t^2 + v0*t
p(t) = (-4.9 m/s^2)*t^2 + v0*t
We know that the maximum height is 2.4m
The value of time at which the fish gets his maximum height is when the velocity of the fish is equal to zero, then we first need to solve:
v(t) = (-9.8m/s^2)*t + v0 = 0
t = v0/9.8m/s^2
Now we replace this in the position equation to get the maxmimum height, which is equal to 2.4m
2.4m = p( v0/9.8m/s^2) = (1/2)*(-9.8 m /s^2)*(v0/9.8m/s^2)^2 + v0*(v0/9.8m/s^2)
2.4m = (1/2)(-v0)^2(-9.8 m /s^2) + v0^2/(9.8m/s^2))
2.4m = (1 - 1/2)*v0^2/(9.8m/s^2)
2.4m = 0.5*v0^2/(9.8m/s^2)
2.4m/0.5 = v0^2/(9.8m/s^2)
4.8m*(9.8m/s^2) = v0^2
√(4.8m*(9.8m/s^2)) = v0 = 6.86 m/s
Calculate the work done to raise a charge of 25 coulombs through an emf of 8 volts.
1) 3
2) 200
Corrected, it's 2) 200
1. (6x + 8)(5x - 8)
a. 30x2 + 49x + 20
2. (5x + 6(5x - 5)
b. 24x3 + 8x2 + 6x + 4
3. (6x + 3)(6x - 4)
c. 25x2 + 5x - 30
4. (6x + 5)(5x + 5)
d. 30x2 - 8x - 64
e. 36x2 - 6x - 1
5. (4x + 2) (6x2 - x + 2)
Answer:
form 1 question??????????
According to Newton’s law of universal gravitation, in which of the following situations does the gravitational attraction between two bodies always increase
Answer:
When the mass increases or when distance between the bodies reduces
Explanation:
According to Newton's law of universal gravitation, the gravitational attraction between two bodies always increase if the mass increases and the distance between the bodies reduces.
The law of universal gravitation states that "the gravitational force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".
Mathematically;
Fg = [tex]\frac{G m1 m2}{r^{2} }[/tex]
G is the universal gravitation constant
m is the mass
r is the distance
Now we’ll use the component method to add two vectors. We will use this technique extensively when we begin to consider how forces act upon an object. Vector A⃗ A→ has a magnitude of 50 cmcm and a direction of 30∘∘, and vector B⃗ B→ has a magnitude of 35 cmcm and a direction of 110∘∘. Both angles are measured counterclockwise from the positive xx axis. Use components to calculate the magnitude and direction of the vector sum (i.e., the resultant) R⃗ =A⃗ +B⃗ R→=A→+B→.
Answer:
The magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°
Explanation:
Since vector A has magnitude 50 cm and a direction of 30, its x - component is A' = 50cos30 = 43.3 cm and its y - component is A" = 50sin30 = 25.
Also, Since vector B has magnitude 35 cm and a direction of 110, its x - component is A' = 35cos110 = -11.97 cm and its y - component is A" = 35sin110 = 32.89 cm.
So, the vector sum R = A + B
The x-component of the vector sum is R' = A'+ B' = 43.3 cm + (-11.97 cm) = 43.3 cm - 11.97 cm = 31.33 cm
The y-component of the vector sum is R" = A"+ B" = 25 cm + 32.89 cm = 57.89 cm
So, the magnitude of R = √(R'² + R"²)
= √((31.33 cm)² + (57.89 cm)²)
= √(981.5689 cm² + 3,351.2521 cm²)
= √(4,332.821 cm²)
= 65.82 cm
≅ 65.8 cm
The direction of R is Ф = tan⁻¹(R"/R')
= tan⁻¹(57.89 cm/31.33 cm)
= tan⁻¹(1.84775)
= 61.58°
≅ 61.6°
So, the magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°
A Geiger–Müller tube can be used to detect ionising radiation. The tube is filled with low pressure gas. How does this detect ionising radiation
Explanation:
the gas will be hit by high energy particles as the radioactive decay happens and the decaying nuclei are breaking apart. these particles energize the gas and cause electrons to move around the orbitals. the flow of electrons is detected by the sensor.
A strong man and a weak man are trying to carry a ladder . How should they carry it in such a way that the weak person feels less weight of the ladder
Answer:
The strong person should carry the ladder at the front end and the weak person should carry it at the back end.
Explanation:
this is because in such a case the strong person has to pull the ladder whereas the weak person at the back end have to push the ladder. In such case it is easier to push because the weak person can use the force of gravity of his own body for pushing the ladde.
However in case of pulling the ladder one has to overcome his own gravity to pull the heavy object
I Hope it help you
Answer:
weak person: near the end of the ladder
strong person: nearer the center of the ladder
Explanation:
The closer the strong person is to the center of mass of the ladder, the greater the fraction of the ladder's weight that person will carry. The weak person will feel less weight if the strong person is nearer to the center of mass.
For example, if the strong person is 3/4 of the ladder's length from the end where the weak person is, then the strong person will carry twice the weight the weak person is having to carry. (Assuming the mass is uniformly distributed.)
How does the force of friction for a sliding object vary with the area of contact?
Answer:
The force and texture of an object matter a lot.
Explanation:
If you were to try and run up a glass hill, could you? You maybe could, but it would be harder to than up carpet.
As you are cycling to the Q center one spring day, you pass the construction site at Gant and stop to watch a few minutes. A crane holds bricks on a pallet to an upper floor of the building. Suddenly, a brick falls off the rising pallet. You clock the time it takes to hit the ground at 2.5 s. From how high did the brick fall?
Answer:
30.63 m
Explanation:
Using y = ut + 1/2gt² where u = initial speed of block = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time of fall = 2.5 s and y = height of fall.
So, substituting the values of the variables into the equation, we have
y = ut + 1/2gt²
y = 0 m/s × 2.5 s + 1/2 × 9.8 m/s² × (2.5 s)²
y = 0 m + 4.9 m/s² × 6.25 s²
y = 0 m + 30.625 m
y = 30.625 m
y ≅ 30.63 m
So, the brick fell 30.63 m
HELP PLS
Which of the following statements about suspensions is TRUE?
a. A suspension contains particles smaller than those of a solution.
b. A suspension cannot be filtered.
c. The large particles in a suspension eventually settle to the bottom of the container.
d. A suspension is a true solution.
Answer:
I think it might be C.
Explanation:
Answer:
c. The large particles in a suspension eventually settle to the bottom of the container.
Explanation:
The correct statement from the given choices is that the large particles in a suspension eventually settle to the bottom of the container.
A suspension is a mixture made up of small insoluble particles of a solid in a liquid or gas.
They have the following properties:
They settle on standingThe particles do not pass through ordinary filter paperThe particles do no pass through permeable membrane Their particles are visible with the naked eyes.You apply a force of 500 N to 150 N/m. how much does it stretch? Show the equation you are using, plus the values into the equation, and show the final answer with the units. 3 significant figures.
Answer:
3.33m
Explanation:
Given parameters:
Force applied =500N
Elastic constant = 150N/m
Unknown:
Amount of stretch or extension = ?
Solution:
To solve this problem use the expression below:
F = k e
F is the force applied
k is the elastic constant
e is the extension
So;
500 = 150 x e
e = [tex]\frac{500}{150}[/tex] = 3.33m
A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from rest and gains a speed of 15 m/s after sliding 150 m. How much work is done against friction
Answer:
[tex]1270.64\ \text{J}[/tex]
Explanation:
m = Mass of object = [tex]\dfrac{mg}{g}[/tex]
mg = Weight of object = 20 N
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
v = Final velocity = 15 m/s
u = Initial velocity = 0
d = Distance moved by the object = 150 m
[tex]\theta[/tex] = Angle of slope = [tex]30^{\circ}[/tex]
f = Force of friction
fd = Work done against friction
The force balance of the system is
[tex]\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}[/tex]
The work done against friction is [tex]1270.64\ \text{J}[/tex].
A 5-kg object is moving with a speed of 4 m/s at a height of 2 m. The potential energy of the object is approximately
J.
Answer:
P.E = 98 Joules
Explanation:
Given the following data;
Mass = 5kg
Speed = 4m/s
Height = 2m
We know that acceleration due to gravity is equal to 9.8m/s²
To find the potential energy;
Potential energy can be defined as an energy possessed by an object or body due to its position.
Mathematically, potential energy is given by the formula;
[tex] P.E = mgh[/tex]
Where, P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
Substituting into the equation, we have;
[tex] P.E = 5*9.8*2[/tex]
P.E = 98 Joules
What is the gravitational potential energy of 4.0 kg pinata suspended 2.5 meters above the ground
Answer:
100 or 95
Explanation:
GPE=M*GFS*change in height
GPE=4.0*10*2.5
=100
For the gravitational field strength you can use 10 or 9.5 but if it says specifically to use a certain value, than use that
Tim has a house worth $250,000. He has installed a $100,000 aquarium in his basement because he is obsessed with fish. His regular homeowner's insurance doesn't cover home aquariums, so his insurance agent is writing a supplemental policy to help him get the coverage for the aquarium. What is this known as? Question 6 options: an exclusion a rider a deductible a consideration
Answer:
A rider.
Explanation:
A rider designates a clause or group of clauses added to an initial contract, being an annex to it. These added clauses are attached to the initial contract, integrating to it and losing the character of an independent document. Therefore, it would then be a piece of writing that is to be treated as part of the main contract. Usually, the purpose of the riders is to include particular clauses for the insured, that is, clauses that are specific to them outside of the company's generic contract.
Answer:
A rider, for plato users
Explanation:
An astronaut with a mass of 75 kg stands on Mars (gravity = 3.7 m/s2). How much does the astronaut weigh?
Answer:
277.5 N.
Explanation:
From the question given above, the following data were obtained:
Mass (m) of astronaut = 75 kg
Acceleration due to gravity on Mar (gₘ) = 3.7 m/s²
Weight of astronaut on Mar (Wₘ) =.?
Weight of an object is simply defined by the following equation:
Weight (W) = mass × acceleration due to gravity (g)
W = m × g
With the above formula, we can obtain the weight of the astronaut on Mar as follow:
Mass (m) of astronaut = 75 kg
Acceleration due to gravity on Mar (gₘ) = 3.7 m/s²
Weight of astronaut on Mar (Wₘ) =.?
Wₘ = m × gₘ
Wₘ = 75 × 3.7
Wₘ = 277.5 N
Thus, the weight of the astronaut on Mar is 277.5 N
We should stress again that the Carnot engine does not exist in real life: It is a purely theoretical device, useful for understanding the limitations of heat engines. Real engines never operate on the Carnot cycle; their efficiency is hence lower than that of the Carnot engine. However, no attempts to build a Carnot engine are being made. Why is that
Answer:
The Carnot engine has zero power
Explanation:
Although theoretically the Carnot engine has more efficiency than the real engine. In practice however they tend to have zero power.
This is because all its processes are reversible (that is isothermic and adiabatic).
So the system equilibrates with its surroundings at every point in time. This makes work done very slow and the power generated is zero.
Carnot cycles requires attaining isothermal heat transfers which is quite difficult and take a long time. Also a pump that can handle liquid-vapour phase mixture will be required.
This is not practical.