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Andrei [34K]
3 years ago
12

How much force is needed to keep a 25 N rock moving with constant speed across the level, frictionless ice?

Physics
1 answer:
jolli1 [7]3 years ago
7 0

Answer:

Explanation:

Hello friend!!!!!

The answer is 0 Newton

hope this helps

plz mark as brainliest!!!!!!!!!!

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If Earth's obliquity was 157 degrees, would the seasons be more severe, less severe, or about the same?
ikadub [295]
The severity of the seasons on Earth is given not by the distance Earth-Sun but by the tilt of the Earth axis. This happens because that the sun rays are oblique in winter and perpendicular in summer (thus the same quantity of sun rays heats a bigger surface in winter - oblique rays). 
The present tild of the Earth axis is  23.5 degrees (from the vertical). If the axis were tilt at 157 degree this would be equivalent  to 180-157 =23 degree. Thus the severity of the seasons would be approximately the same but the seasons would be reversed (for example instead of winter we would have summer, instead of summer we would have winter). 
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3 years ago
The ability of solid rock to flow is called _____.
zysi [14]
The ability of solid rock to flow is called Plasticity.

Plasticity is the deformation of a solid mateial which resulted in non reversible changes of shape in response to applied force.

Another example of plasticity is when you bend metal through a forcer force to create an art or kitchen set
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3 years ago
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Which of the following is NOT a function of the lens in the eye?
tatyana61 [14]

Answer:

it is a light receptor that generates nerve signals that are sent to the brain

Explanation:

the lens are like the glasses, this means that is used to see things better. You just put them in your eye and that's all it's not connected to the brain

8 0
2 years ago
A box is dropped onto a conveyor belt moving at 3.2 m/s. If the coefficient of friction between the box and the belt is 0.28, ho
Lemur [1.5K]

Answer:

t = 1.16 s.

Explanation:

Given,

speed of conveyor belt, v = 3.2 m/s

coefficient of friction,f = 0.28

Using newton second law

f = ma

and we also know that frictional force

f = μ N

f = μ m g

equating both the force equation

a = μ g

a = 0.28 x 9.81

a = 2.75 m/s²

Using Kinematic equation

v = u + at

3.2 = 0 + 2.75 x t

t = 1.16 s.

Time taken by the box to move without slipping is 1.16 s.

6 0
2 years ago
A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .
rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

v=u+at

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

v = 15.5 +(-9.8)(2.64)=-10.4 m/s

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

5 0
3 years ago
Read 2 more answers
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