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3 years ago

I need the ans for this question QUICK PLEASE!!!

Physics

1 answer:

Basile [38]3 years ago

3 0

Explanation:

A) 1.05

B) 1.33

C) 1.16

D) 0.62

All units in cm

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The Hubble telescope’s orbit is 5.6 × 105 meters above Earth’s surface. The telescope has a mass of 1.1 × 104 kilograms. Earth e

djverab [1.8K]

The gravitational field is the Force divided by the mass

Call g the gravitational fiel, F the force exerted by the earth and m the mass of the telescope.

g = F / m

g=9.1x10^4 N / 1.1 x 10^4 kg = 8.27 N/kg

Note that the unit N/kg is equivalent to m/s^2

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4 years ago

When water fills a submarine’s flotation tanks, what happens to the submarine?

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Answer:

It over all density increases and it begins to sink.

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3 years ago

A 0.2 kg baseball is pitched with a velocity of 40 m/s and is then batted to the pitcher with a velocity of 60 m/s. What is the

sergey [27]

Answer:

The magnitude of change in the ball's momentum is 4 kgm/s

Explanation:

Given;

mass of the ball, m = 0.2 kg

initial velocity of the ball, u = 40 m/s

final velocity experienced by the ball, v = 60 m/s

Therefore, the change in momentum of the ball is given as final momentum minus initial mometum;

ΔP = mv - mu

ΔP = m(v-u)

ΔP = 0.2 (60 - 40)

ΔP = 4 kgm/s

Therefore, the magnitude of change in the ball's momentum is 4 kgm/s

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3 years ago

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notka56 [123]

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4 years ago

The body falls in a free fall 11s.Calculate: a) from what height the body of the paddle) what path did it fall during the last s

JulsSmile [24]

Answer:

a) h = 593.50 m

b) h₁₁ = 103 m

c) vf = 107.91 m/s

Explanation:

We will use second equation of motion to find the height:

$h = v_{i}t + (\frac{1}{2})gt^2$

where,

h = height = ?

vi = initial speed = 0 m/s

t = time taken = 11 s

g = 9.81 /s²

Therefore,

$h = (0\ m/s)(11\ s) + (\frac{1}{2})(9.8\ m/s^2)(11\ s)^2\\\\$

h = 593.50 m

For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:

$v_{f} = v_{i} + gt$

where,

vf = final velocity at tenth second = v₁₀ = ?

t = 10 s

vi = 0 m/s

Therefore,

$v_{10} = 0\ m/s + (9.81\ m/s^2)(10\ s)\\\\v_{10} = 98.1\ m/s$

Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:

$h = v_{i}t + (\frac{1}{2})gt^2$

where,

h = height covered during last second = h₁₁ = ?

vi = v₁₀ = 98.1 m/s

t = 1 s

Therefore,

$h_{11} = (98.1\ m/s)(1\ s) + (\frac{1}{2})(9.8\ m/s^2)(1\ s)^2\\\\$

h₁₁ = 103 m

Now, we use first equation of motion for complete motion:

$v_{f} = v_{i} + gt$

where,

vf = final velocity at tenth second = ?

t = 11 s

vi = 0 m/s

Therefore,

$v_{f} = 0\ m/s + (9.81\ m/s^2)(11\ s)$

vf = 107.91 m/s

8 0

3 years ago