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3 years ago

A rock on the ground weighs 40N. How much force does the ground exert on the rock?

Physics

1 answer:

worty [1.4K]3 years ago

5 0

Answer:

The correct option is;

C. 40 N

Explanation:

Given that the rock is on the ground, we have;

The weight of the rock, $F_W$ = 40 N ↓

From Newton's third law of motion, we have that action and reaction are equal and opposite,

Therefore, given that the weight of the rock exerts a force of 40 N on the ground, and the rock is stable and stationery on the ground, then the ground is reacting to the weight of the rock with an equal force of 40 N, acting in opposite direction to the weight of the block to balance the weight of the rock and create equilibrium such that the net force acting on the rock is zero

∑F = $F_W + F_R$ = 0

Where;

$F_R$ = The reaction of the ground

We have;

$F_W$ + $F_R$ = 40 + $F_R$ = 0

$F_R$ = 0 - 40 = -40

$F_R$ = -40 N ↓

∴ $F_R$ = 40 N ↑.

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A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

4 0

3 years ago

A body is dropped from the roof of a 20 m high building by how much:

USPshnik [31]

Answer:

t = 2.01 s

Vf = 19.7 m/s

Explanation:

It's know through the International System that the earth's gravity is 9.8 m/s², then we have;

Data:

Height (h) = 20 m
Gravity (g) = 9.8 m/s²
Time (t) = ?
Final Velocity (Vf) = ?

==================================================================

Time

Use formula:

$\boxed{t=\sqrt{\frac{2*h}{g}}}$

Replace:

$\boxed{t=\sqrt{\frac{2*20m}{9.8\frac{m}{s^{2}}}}}$

Everything inside the root is solved first. So, we solve the multiplication of the numerator:

$\boxed{t=\sqrt{\frac{40m}{9.8\frac{m}{s^{2}}}}}$

It divides:

$\boxed{t=\sqrt{4.08s}}$

The square root is performed:

$\boxed{t=2.01s}$

==================================================================

Final Velocity

use formula:

Vf = g * t

Replace:

Vf = 9.8 m/s² * 2.01 s

Multiply:

Vf = 19.7 m/s

==================================================================

How long does it take to reach the ground?

Takes time to reach the ground in <u>2.01 seconds.</u>

How fast does it hit the ground?

Hits the ground with a speed of <u>19.7 meters per seconds.</u>

7 0

3 years ago

What is the ratios of sodium hydroxide

Katena32 [7]

Clarify what you mean by ratios?

8 0

3 years ago

What is the significance of Nucleotides in Chromosomes?

fgiga [73]

Answer:

it comprises of the DNA/RNA bipolymer molecules

3 0

3 years ago

Read 2 more answers

Can a body have zero velocity and finite acceleration?Explain

sergejj [24]

Answer:

Kinda? Depends what the question is fully asking

Explanation:

Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.

So that -2 m/s thing after one second will be going -1 m/s.

After another second it'll be going 0 m/s.

After another itll be going +1 m/s and so on.

So at one point for a brief moment, it can have an acceleration but be at 0 m/s velocity.

5 0

3 years ago