All categories

3 years ago

Rock can melt at a depth of about

Physics

2 answers:

Assoli18 [71]3 years ago

6 0

The answer is D 2000 km..

Fed [463]3 years ago

3 0

Answer:

The answer is D, around 2000 km

Explanation:

You might be interested in

9. A plane starts at rest & accelerates along the ground before takeoff. It

Phoenix [80]

Answer:

9.877 m/s^2

Explanation:

The acceleration can be computed from ...

d = (1/2)at^2

(1600 m) = (1/2)a(18 s)^2

a = (1600/162) m/s^2 ≈ 9.877 m/s^2

6 0

4 years ago

2) A skier stands at rest and begins to ski downhill with an acceleration of 3.0 m/s² {downhill). What is

olga2289 [7]

Answer:

her displacement <em>s=337.5m</em>

Explanation:

check out the above attachment ☝️

7 0

2 years ago

Read 2 more answers

An object has a 600 N force pushing it to the right, while a 700 N force pushes in the opposite direction. Gravity also acts on

nordsb [41]

Left is the answer 3

6 0

3 years ago

Read 2 more answers

A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cable

kompoz [17]

Answer:

4.44s

Explanation:

A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables

since the mass of the child and that of the swing is negligible, the masses wont be involved in the calculation

T=2π√L/g

g=acceleration due to gravity which is 9.81m/s2

the length of the supporting cable is 4.9m

T the period

period is the time required to make a complete oscillation

T=2*π√4.9/9.81

T=2*π*0.706

T=4.44s

4.44s

5 0

4 years ago

A 5kg bag falls a verticle height of 10m before hitting the ground.

g100num [7]

Answer:

$u = 7m {s}^{ - 1}$

Explanation:

We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:

$me(1) = me(2)$

where me(1) is mechanical energy while on h=10m

and me(2) is mechanical energy while on the ground

Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)

Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.

DynamicE(2) is equal to zero since it's touching the ground

Using that info we have

$m \times g \times h = \frac{1}{2} \times m \times u {}^{2} \\$

we divide both sides of the equation with mass to make the math easier.

$9.8 \times 10 = \frac{1}{2} \times u {}^{2} \\ \frac{98}{2} = u {}^{2} \\ u { }^{2} = 49 \\ u = 7$

7 0

4 years ago