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Fofino [41]
3 years ago
13

What is the unique geological feature found on Mercury surface?

Physics
1 answer:
Cloud [144]3 years ago
3 0

Answer:

The surface of Mercury has landforms that indicate its crust may have contracted. They are long, sinuous cliffs called lobate scarps. These scarps appear to be the surface expression of thrust faults, where the crust is broken along an inclined plane and pushed upward.

Explanation:

I hope this helps a little bit.

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what is the approximate weight of a 20-kg cannonball on the moon if the acceleration due to gravity is 1.6m/s^2
monitta
On Earth, a cannonball with a mass of 20 kg would weigh 196 Newtons.
With the formula F=mg, where F is the weight in Newtons, m is the mass, and g is the acceleration due to gravity on the Earth which is 9.8m/s^2.
F=20kg x 9.8m/s^2= 196 Newtons

BUT on the moon, acceleration due to gravity is 1.6 m/s^2,
so F=mg=20kgx1.6m/s^2= 32 N
5 0
3 years ago
11. A 3.8 kg object is lifted 12 meters. Approximately how much work is performed during the lifting?
Marianna [84]
I found the answer for you if u need any help ask anytime!

3 0
3 years ago
A sprinter with a mass of 80 kg accelerates from 0 m/s to 9 m/s in 3 s. What is the runner's acceleration?
sertanlavr [38]
Acceleration = (change in speed) / (time for the change) = 9/3 = <em>3 m/s²</em> .

His mass makes no difference.
5 0
3 years ago
Which two options are examples of waves reflecting?
Zinaida [17]
I think the answer is b
3 0
2 years ago
Read 2 more answers
In our first example we will consider a very simple application of Newton’s second law. A worker with spikes on his shoes pulls
sweet-ann [11.9K]

Answer:

Acceleration=0.5m/s^2

Speed=0.67 m/s

Explanation:

We are given that

Horizontal force=F=20 N

Mass of box=m=40 kg

We know that

Acceleration=a=\frac{F}{m}

Using the formula

Acceleration of box=\frac{20}{40}=0.5m/s^2

The acceleration of the box=0.5m/s^2

Initial velocity=u=0

Force=F=30 N

Distance=s=0.3 m

a=\frac{30}{40}=\frac{3}{4} ms^{-2}

v^2-u^2=2as

Substitute the values

v^2-0=2\times \frac{3}{4}\times 0.3=0.45

v^2=0.45

v=\sqrt{0.45}=0.67m/s

Hence, the speed of the box after it has  been pulled a distance of 0.3 m=0.67 m/s

4 0
3 years ago
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