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3 years ago

10

A worker does 25 J of work lifting a bucket, then sets the bucket back down in the same place. What is the total net work done o

n the bucket?

1 answer:

Vera_Pavlovna [14]3 years ago

5 0

Answer:

0 J

Explanation:

As work is force times displacement, if no displacement occurs, no work occurs.

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A ball is thrown with velocity of 10 m/s upwards. If the ball is caught 1 m above its initial position, what is the speed of the

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Answer:

v = 8.96 m/s

Explanation:

Initial speed of the ball, u = 10 m/s

It caught 1 meter above its initial position.

Acceleration due to gravity, $g=-9.8\ m/s^2$

We need to find the final speed of the ball when it is caught. Let is equal to v. To find the value of v, use third equation of motion as :

$v^2-u^2=2as$

$v^2=2as+u^2$

$v^2=2(-9.8)\times 1+(10)^2$

v = 8.96 m/s

So, the speed of the ball when it is caught is 8.96 m/s. Hence, this is the required solution.

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Problem 2: A rock is dropped from a cliff. If it falls for 10 seconds, what is its displacement?

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A solid sphere of uniform density has a mass of 3.0 × 104 kg and a radius of 1.0 m. What is the magnitude of the gravitational f

elena55 [62]

Answer:

a) Fg = 9.495x10⁻⁶N

b) Fg = 3.908x10⁻⁶N

c)

$F_{g} =\frac{Gm_{1}m_{2}R }{r^{3} }$

Explanation:

Given:

m₁ = mass = 3x10⁴kg

r = radius = 1 m

m₂ = 9.3 kg

Questions:

a) What is the magnitude of the gravitational force due to the sphere located at R = 1.4 m, Fg = ?

b) What is the magnitude of the gravitational force due to the sphere located at R= 0.21 m, Fg = ?

c) Write a general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.0 m from the center of the sphere.

a) Since R > r, the equation for the gravitational force is:

$F_{g} =\frac{Gm_{1}m_{2} }{R^{2} }$

Here,

G = gravitational constant = 6.67x10⁻¹¹m³/s² kg

Substituting values:

$F_{g} =\frac{6.67x10^{-11}*3x10^{4}*9.3 }{1.4^{2} } =9.495x10^{-6} N$

b) Since R < r, the equation for the gravitational force is:

$F_{g} =\frac{Gm_{1}m_{2}R }{r^{3} } =\frac{6.67x10^{-11}*3x10^{4}*9.3*0.21 }{1^{3} } =3.908x10^{-6} N$

c) The general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.0 is the same to b)

$F_{g} =\frac{Gm_{1}m_{2} R}{r^{3} }$

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