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3 years ago

7

Gravity pulls downward on a rock with a force of 800 N. If you pull upward on the rock with a force of 400 N, what is the total

force acting on the rock?
A. 400 N downward
B. 400 N upward
C. 1200 N downward
D. 1200 N upward

2 answers:

aleksley [76]3 years ago

6 0

A i believe srry if wrong

wel3 years ago

3 0

The answer is
a. 400 N downward

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A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to

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Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

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2 years ago

Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters

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Answer:

C. $\frac{3F}{8}$

Explanation:

Let initial charges on both spheres be, $q$

$F=\frac{Kq^2}{d^2} \ \ \ \ \ \ \ \ \ \ \_i$

When the sphere C is touched by A, the final charges on both will be, $\frac{q}{2}$

#Now, when C is touched by B, the final charges on both of them will be:

$q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\$

Now the force between A and B is calculated as:

$F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}$

Hence the electrostatic force becomes 3F/8

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2 years ago

What is the most appropriate tool to use when collecting data from the investigation? List, table, graph or chart?

forsale [732]

A table would be the most appropriate because that way you can compare the data.

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2 years ago

Assuming a 8 kilogram bowling ball moving at 2 m/s bounces off a spring at the same speed that had before bouncing what is the a

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a) 32 kg m/s

Assuming the spring is initially at rest, the total momentum of the system before the collision is given only by the momentum of the bowling ball:

$p_i = m u = (8 kg)(2 m/s)=16 kg m/s$

The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

$p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s$

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

$p_f = p_{fB}+p_{fS}$

where $p_{fS}$ is the momentum of the spring. For the conservation of momentum,

$p_i = p_f\\p_i = p_{fB}+p_{fS}\\p_{fS}=p_i -p_{fB}=16 kg m/s -(-16 kg m/s)=32 kg m/s$

b) -32 kg m/s

The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:

$\Delta p=p_{fb}-p_i=-16 kg m/s - 16 kg m/s=-32 kg m/s$

c) 64 N

The change in momentum is equal to the product between the average force and the time of the interaction:

$\Delta p=F \Delta t$

Since we know $\Delta t=0.5 s$ , we can find the magnitude of the force:

$F=\frac{\Delta p}{\Delta t}=\frac{-32 kg m/s}{0.5 s}=-64 N$

The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.

d) The force calculated in the previous step (64 N) is larger than the force of 32 N.

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