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3 years ago

A horse pulls on a wagon. The reaction force is the wagon pulls on the horse. true or false?

Physics

2 answers:

ad-work [718]3 years ago

8 0

True because the horses are causing it so do it and it’s moving out of reaction

Katen [24]3 years ago

6 0

Answer:

True

Explanation:

the horse pull on the wagon but friction and the wight + gravity make the wagon pull on the horse ( newton's 2ed law)

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A mass with a charge of 4.60 x 10-7 C rests on a frictionless surface. A compressed spring exerts a force on the mass on the lef

stira [4]

Answer:

The compression of the spring is 24.6 cm

Explanation:

magnitude of the charge on the left, q₁ = 4.6 x 10⁻⁷ C

magnitude of the charge on the right, q₂ = 7.5 x 10⁻⁷ C

distance between the two charges, r = 3 cm = 0.03 m

spring constant, k = 14 N/m

The attractive force between the two charges is calculated using Coulomb's law;

$F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(4.6\times 10^{-7})(7.5\times 10^{-7})}{(0.03)^2} \\\\F= 3.45 \ N$

The extension of the spring is calculated as follows;

F = kx

x = F/k

x = 3.45 / 14

x = 0.246 m

x = 24.6 cm

The compression of the spring is 24.6 cm

4 0

3 years ago

1. A 455 g mass, hanging at rest on a spring, stretches the spring 22.4 cm beyond its relaxed

pishuonlain [190]

Answer:

19.9 N/m

Explanation:

From the question,

Applying Hook's law

F = Ke.................. Equation 1

Where F = Force on the spring, k = spring constant, e = extension

But the force on the spring is the weight of the mass

Therefore,

mg = ke.................. Equation 2

Where m = mass. g = acceleration due to gravity

make e the subject of the equation

e = mg/e................ Equation 3

Given: m = 455 g = 0.455 kg, e = 22.4 cm = 0.224 m,

Constant: g = 9.8 m/s²

Substitute these values into equation 3

e = (0.455×9.8)/0.224

e = 19.9 N/m

6 0

3 years ago

Starting from rest, a runner reaches a speed of 2.8 m/s in 2.1 s. In the same time 2.1 s time, a motorcycles increases speed fro

maw [93]

Answer:

a) Acceleration of runner is 1.33 m/s²

b) Acceleration of motorcycle is 2.85 m/s²

c) The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{2.8-0}{2.1}\\\Rightarrow a=1.33\ m/s^2$

Acceleration of runner is 1.33 m/s²

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{43-37}{2.1}\\\Rightarrow a=2.85\ m/s^2$

Acceleration of motorcycle is 2.85 m/s²

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{2.8^2-0^2}{2\times 1.33}\\\Rightarrow s=2.94\ m$

The runner moves 2.94 m

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{43^2-37^2}{2\times 2.85}\\\Rightarrow s=84.21\ m$

The motorcycle moves 84.21 m

The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.

3 0

3 years ago

What is a Super Massive Black Hole?

Serggg [28]

Answer:

I dont really know much but i know that it swallow anything it comes across in space.

4 0

3 years ago

Two passenger trains are passing each other on adjacent tracks. train a is moving east with a speed of 13 m/s, and train b is tr

alukav5142 [94]

Since the two trains are passing in opposite directions, so this means that their relative velocities will be the sum of the two trains that is:

relative velocity = (13 + 28) = 41m/s

a. The passengers aboard on train B will see that train A is moving at 41m/sec due east

b. The passengers aboard on train A will see that train B is moving at 41m/sec due west

7 0

4 years ago

Read 2 more answers