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2 years ago

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An upward force is applied to a 6.0–kilogram box. This force displaces the box upward by 10.00 meters. What is the work done by

the force on the box?
a. 6.0 × 101 joules
b. -6.0 × 101 joules
c. 5.9 × 102 joules
d. -5.9 × 102 joules

1 answer:

devlian [24]2 years ago

3 0

Answer:

W ≈ 5.9 × $10^{2}$ J

Hence, option (c) is correct.

Given:

mass of box = 6 kg

Displacement of box = 10 m

To find:

Work done by the force = ?

Formula used:

Work done is equal to change is potential energy,

W = m g h

Where, W = Work done

m = mass of the box

h = displacement by the force

Solution:

Work done is equal to change is potential energy,

W = m g h

Where, W = Work done

m = mass of the box

h = displacement by the force

W = 6 × 9.8 ×10

W = 588 Joule

Thus, W ≈ 5.9 × $10^{2}$ J

Hence, option (c) is correct.

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For a metal that has an electrical conductivity of 7.1 x 107 (Ω-m)-1, do the following: (a) Calculate the resistance (in Ω) of a

jonny [76]

Answer:

(a) 0.0178 Ω

(b) 3.4 A

(c) 6.4 x 10⁵ A/m²

(d) 9.01 x 10⁻³ V/m

Explanation:

(a)

σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹

d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) π d²

A = (0.25) (3.14) (2.6 x 10⁻³)²

A = 5.3 x 10⁻⁶ m²

L = length of the wire = 6.7 m

Resistance of the wire is given as

$R=\frac{L}{A\sigma }$

$R=\frac{6.7}{(5.3\times10^{-6})(7.1\times10^{7}) }$

R = 0.0178 Ω

(b)

V = potential drop across the ends of wire = 0.060 volts

i = current flowing in the wire

Using ohm's law, current flowing is given as

$i = \frac{V}{R}$

$i = \frac{0.060}{0.0178}$

i = 3.4 A

(c)

Current density is given as

$J = \frac{i}{A}$

$J = \frac{3.4}{5.3\times10^{-6}}$

J = 6.4 x 10⁵ A/m²

(d)

Magnitude of electric field is given as

$E = \frac{J}{\sigma }$

$E = \frac{6.4 \times 10^{5}}{ 7.1 \times 10^{7}}$

E = 9.01 x 10⁻³ V/m

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2 years ago

A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi

OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn = normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

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2 years ago

A carousel - a horizontal rotating platform - of radius r is initially at rest, and then begins to accelerate constantly until i

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Answer:

α = (ω²)/8π

Explanation:

The angular acceleration(α) of the carousel can be determined by using rotational kinematics:

ω² =ωo² + 2αθ

Let's make α the subject of this equation ;

ω² - ωo² = 2αθ

α = (ω² −ωo²)/2θ

Now, from the question, since initially at rest, thus, ωo = 0

Also,since 2 revolutions, thus, θ = 2 x 2π = 4π since one revolution is 2π

Plugging in the relevant values to get ;

α = (ω²)/2(4π)

α = (ω²)/8π

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2 years ago

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

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Which statements describe vectors? Check all that apply. -Vectors have magnitude only. -Vectors have direction only. -Vectors ha

Natali [406]

Answer:

Vectors have both magnitude and direction

Explanation:

Vectors show how strong the force in because the bigger the arrow, the stronger the force. Also, it obviously shows direction because its an arrow.

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