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4 years ago

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Physics

1 answer:

BlackZzzverrR [31]4 years ago

3 0

In Newton's third law, the action and reaction forces D.)act on different objects

Explanation:

Newton's third law of motion states that:

<em>"When an object A exerts a force on object B (action force), then action B exerts an equal and opposite force (reaction force) on object A"</em>

It is important to note from the statement above that the action force and the reaction force always act on different objects. Let's take an example: a man pushing a box. We have:

Action force: the force applied by the man on the box, forward
Reaction force: the force applied by the box on the man, backward

As we can see from this example, the action force is applied on the box, while the reaction force is applied on the man: this means that the two forces do not act on the same object. This implies that whenever we draw the free-body diagram of the forces acting on an object, the action and reaction forces never appear in the same diagram, since they act on different objects.

Learn more about Newton's third law of motion:

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If a resultant vector is 12 m/s and the horizontal component is 9 m/s, what is the value of the vertical component? A. 5.8 m/s B

horrorfan [7]

The vertical component is B) 7.9 m/s

Explanation:

A vector can be resolved into its horizontal and vertical components. The horizontal and the vertical components form the sides of a right triangle, in which the resultant corresponds to the hypothenuse, so we can use Pythagorean's theorem:

$R=\sqrt{R_x^2+R_y^2}$

where

R is the resultant vector

Rx is the horizontal component

Ry is the vertical component

In this problem, we have:

R = 12 m/s is the resultant vector

Rx = 9 m/s is the horizontal component

Solving the formula for Ry, we find the vertical component:

$R_y = \sqrt{R^2-R_x^2}=\sqrt{12^2-9^2}=7.9 m/s$

Learn more about vector components:

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3 0

3 years ago

A 23.3 kg crate is pushed with a force of 944 N for 36.0 seconds, moving the crate a distance of 12.4 m. How much power was used

tigry1 [53]

Explanation:

$p = \frac{f \times s}{t}$

power = Force × distance /time

power = 944N × 12.4m/36secs

power = (944×12.4/36)Nms—¹

power = 390.2Nms—¹ or 390.2Watts or 390.2Js—¹

8 0

3 years ago

We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm . The cylinder rotates with

amm1812

Answer: 14.16

Explanation:

Given

d = 38cm

r = d/2 = 38/2 = 19cm = 0.19m

K.E = 510J

m = 10kg

I = 1/2mr²

I = 1/2*10*0.19²

I = 0.18kgm²

When it has 510J of Kinetic Energy then,

510J = 1/2Iω²

ω² = 1020/I

ω² = 1020/0.18

ω² = 5666.67

ω = √5666.67 = 75.28 rad/s

Velocity is the block, v = ωr

V = 75.28 * 0.19

V = 14.30m/s

The "effective mass" M of the system is

M = (14.0 + ½*10.0) kg = 19.0 kg

The motive force would be

F = ma

F = 14 * 9.8

F = 137.2N

so that the acceleration would be

a = F/m

a = 137.2/19

a = 7.22m/s²

Finally, using equation of motion.

V² = u² + 2as

14.3² = 0 + 2*7.22*s

204.49 = 14.44s

s = 204.49/14.44

s = 14.16m

6 0

3 years ago

A 13.0 g piece of Styrofoam carries a net charge of â5.40 Î¼C and floats above the center of a large horizontal sheet of plastic

kati45 [8]

Answer:

σ =4.180×10^{-9} C/m^2

Explanation:

electric field due to non conducting sheet is

$E=\frac{\sigma}{2\epsilon_0}$

the force acting on the piece of Styrofoam

Eq= mg

⇒E= mg/q

now,

$\frac{mg}{q} = \frac{\sigma}{2\epsilon_0}$

⇒ $\sigma= \frac{2mg\epsilon_0}{q}$

$\sigma= \frac{2\times0.013\times9.81\times8.85\times10^{-12}}{5.40\times10^{-4}}$

charge per unit area (in C/m2) on the plastic sheet σ =4.180×10^{-9} C/m^2

6 0

4 years ago

A 1500 kg car moving with a speed of 20 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of

storchak [24]

Answer:

-100000 N.

Explanation:

Force: This can be defined as the product of the mass of a body and it's acceleration. The S.I unit of Force is Newton(N). The Formula of force is given as,

F = ma ........................... Equation 1

Where F = Average force exerted on the car, m = mass of the car, acceleration of the car, a = acceleration of the car.

a = (v-u)/t..................... Equation 2

Where v = Final velocity, u = Initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t............. Equation 3

Given: m = 1500 kg, u = 20 m/s, v = 0 m/s (brought to rest), t = 0.3 s.

Substitute into equation 3

F = 1500(0-20)/0.3

F = 1500(-20)/0.3

F = -100000 N.

Note: The negative sign is due to the fact that the force exerted on the car by the pole is equal and opposite the force of the car.

7 0

4 years ago