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3 years ago

A force of 42N is needed to start a box sliding across the floor. The weight of the box is 55N.

Physics

2 answers:

Bezzdna [24]3 years ago

4 0

I think the right answer is c

GenaCL600 [577]3 years ago

3 0

Explanation:

It is given that,

Force applied to the box, F = 42 N

Weight of the box, W = 55 N

(a) Let f is the force of the friction acting on the box. The force of friction always opposes the motion of an object. It always equal to the applied force as :

f = -F = -42 N

(b) The frictional force is kinetic in nature because when the force is applied to the box it starts to move.

(c) Let $\mu$ is the coefficient of friction. It is given by :

$\mu=\dfrac{f}{N}$

$\mu=\dfrac{42}{55}$

$\mu=0.76$

Hence, this is the required solution.

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A V = 108-V source is connected in series with an R = 1.1-kΩ resistor and an L = 34-H inductor and the current is allowed to rea

soldi70 [24.7K]

Answer:

Explanation:

Given an RL circuit

A voltage source of.

V = 108V

A resistor of resistance

R = 1.1-kΩ = 1100 Ω

And inductor of inductance

L = 34 H

After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor

A. Time the inductor current will reduce to 12% of it's initial current

Let the initial charge current be Io

Then, final current is

I = 12% of Io

I = 0.12Io

I / Io = 0.12

The current in an inductor RL circuit is given as

I = Io ( 1—exp(-t/τ)

Where τ is time constant and it is given as

τ = L/R = 34/1100 = 0.03091A

So,

I = Io ( 1—exp(-t/τ))

I / Io = ( 1—exp(-t/τ))

Where I/Io = 0.12

0.12 = 1—exp(-t/τ)

0.12 — 1 = —exp(-t/τ)

-0.88 = -exp(-t/0.03091)

0.88 = exp(-t/0.03091)

Take In of both sides

In(0.88) = In(exp(-t/0.03091)

-0.12783 = -t/0.030901

t = -0.12783 × 0.030901

t = 3.95 × 10^-3 seconds

t = 3.95 ms

B. Energy stored in inductor is given as

U = ½Li²

So, the current at this time t = 3.95ms

I = Io ( 1—exp(-t/τ))

Where Io = V/R

Io = 108/1100 = 0.0982 A

Now,

I = Io ( 1—exp(-t/τ))

I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))

I = 0.0982(1—exp(-0.12783)

I = 0.0982 × 0.12

I = 0.01178

I = 11.78mA

Therefore,

U = ½Li²

U = ½ × 34 × 0.01178²

U = 2.36 × 10^-3 J

U = 2.36 mJ

8 0

3 years ago

Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

kramer

Answer:

$v_x = -6\ m/s$

Explanation:

initial velocity

magnitude of velocity, v = 12 m/s

angle made of velocity with negative x-axis,θ = 60°

We need to calculate x- component of v

$v_x = v cos \theta$

velocity is in negative x-direction, v = -12 m/s

now,

$v_x = -12\times cos 60^0$

$v_x = -12\times 0.5$

$v_x = -6\ m/s$

Hence, the velocity x-component is equal to -6 m/s.

5 0

3 years ago

A 24 kg child slides down a 3.3-m-high playground slide. She starts from rest, and her speed at the bottom is 3.0 m/s.a. What en

Gelneren [198K]

Answer:

(a) Potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

(b) $U=668.16\ J$

Explanation:

Given:

mass of the child, $m=24\ kg$

height of the slide, $h=3.3\ m$

initial velocity of the child at the slide, $v_i=0 m.s^{-1}$

final velocity of the child at the bottom of slide, $v_f=3\ m.s^{-1}$

(a)

∴The initial potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

Initial potential energy:

$PE=m.g.h$