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3 years ago

11. A depression in the ground due to a cave collapse or acidic water dissolution of limestone is called a

Physics

1 answer:

Ierofanga [76]3 years ago

4 0

It is a sinkhole so yeah.. letter A

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A car is moving from rest and the velocity increases to 30 m/s in 4 seconds. Calculate its acceleration.

vaieri [72.5K]

Answer:

7.5 m/s²

Explanation:

Given:

v₀ = 0 m/s

v = 30 m/s

t = 4 s

Find: a

v = at + v₀

(30 m/s) = a (4 s) + (0 m/s)

a = 7.5 m/s²

8 0

3 years ago

If a 2,000-kg car hits a tree with 500 n of force over a time of 0.5 seconds, what is the magnitude of its impulse?

Elden [556K]

Impulse = Force * time
Impulse = 500N *0.5 s =250 N*s

4 0

3 years ago

Read 2 more answers

The distance traveled by an object can be modeled by the equation d = ut + 0.5at2 where d = distance, u = initial velocity, t =

Taya2010 [7]

We are given with the expression d = ut + 0.5 at^2 and is asked to express the equation in terms of a. First, we transpose ut to the left side, then we multiply to the equation and divide lastly the resulting equation by t^2. The final expression becomes a = 2(d-ut)/t^2.

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3 years ago

How do Newton’s laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can pass a pu

notka56 [123]

Answer:

Newton's first law

Explanation:

Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. Therefore, when the ice is smooth, friction gets lesser, and the force acted on that Puck will be decreased.

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3 years ago

Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a

suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

$F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N$

Part (b)

the magnitude of each charge, if they have equal magnitude

$F = \frac{KQ^2}{r^2}$

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

$F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q = \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C$

4 0

3 years ago