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Lemur [1.5K]
2 years ago
11

How high is an object lifted if the total work done on it is 125 J and the force required to lift the object is 25 N? Use the eq

uation W = F x d.
A. 6 m
B. 12 m
C. 5 m
D. 10 m
Physics
1 answer:
mixas84 [53]2 years ago
8 0
The answer would be C. 5m
This is because to find d, you would need to divide W (125 J) by F (25 N).
Hope this helps!
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Which of the following correctly shows the chain of energy transfers that create surface currents on the ocean?
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D.
Solar energy is converted to wind energy which then drive surface currents.
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Water is different from other substances because:a it is more dense as a solid than a liquidb it is less dense as a solid than l
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B is correct. Ever wondered why ice floats in water?
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The fast train known as the TGV (Train à Grande Vitesse) that runs south from Paris, France, has a scheduled average speed of 21
andrezito [222]

Answer:

Explanation:

Given

average speed of train(v_{avg})=216 kmph\approx 60 m/s

Maximum acceleration=0.05g

Now centripetal acceleration is

a_c=\frac{v^2}{r}

0.05\times 9.8=\frac{60^2}{r}

r=7346.93 m

(b)Radius of curvature=900 m

therefore a_c=\frac{v^2}{r}

v=\sqrt{a_cr}

v=\sqrt{0.05\times 9.8\times 900}

v=\sqrt{441}=21 m/s

8 0
3 years ago
A 1,000 kg car is travelling at 6.5 m/s to the North. A 3,500 kg truck is travelling South at the same velocity. What is the tot
Molodets [167]

Answer:

16250 kgm/s due south

Explanation:

Applying,

M = mv................. Equation 1

Where M = momentum, m = mass, v = velocity.

From the car,

Given: m = 1000 kg, v = 6.5 m/s

Substitute these values into equation 1

M = 1000(6.5)

M = 6500 kgm/s

For the truck,

Given: m = 3500 kg, v = 6.5 m/s

Substitute these values into equation 1

M' = 3500(6.5)

M' = 22750 kgm/s.

Assuming South to be negative direction,

From the question,

Total momentum of the two vehicles = (6500-22750)

Total momentum of the two vehicles = -16250 kgm/s

Hence the total momentum of the two vehicles is 16250 kgm/s due south

3 0
2 years ago
A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik
shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle=40^{\circ}

x =12\pm 0.27

y=1.05

equation of trajectory of ball is given by

y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }

for x=12.27

1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }

u=10.69

for x=11.73

1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }

u=11.436 m/s

Thus range of speed is (10.69, 11.436)

3 0
2 years ago
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