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2 years ago

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30) A force produces power P by doing work W in a time T. What power will be produced by a force that does six times as much wor

k in half as much time?
A) 12P
B) 6P
C) P
D) P
E) P

1 answer:

schepotkina [342]2 years ago

4 0

Answer:

A) 12P

Explanation:

The power produced by a force is given by the equation

$P=\frac{W}{T}$

where

W is the work done by the force

T is the time in which the work is done

At the beginning in this problem, we have:

W = work done by the force

T = time taken

So the power produced is

$P=\frac{W}{T}$

Later, the force does six times more work, so the work done now is

$W'=6W$

And this work is done in half the time, so the new time is

$T'=\frac{T}{2}$

Substituting into the equation of the power, we find the new power produced:

$P'=\frac{W'}{T'}=\frac{6W}{T/2}=12\frac{W}{T}=12P$

So, 12 times more power.

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A ship sailing in the Gulf Stream is heading 25.0º west of north at a speed of 4.00 m/s relative to the water. Its velocity rela

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Answer:

Velocity of Gulf Stream= $34.5^\circ$ west of north at a speed of $2.03\ \rm m/s$

Explanation:

Given

speed of ship relative to Gulf stream= $25^\circ$ west of north at a speed of 4 m/s

speed of ship relative to earth= $5^\circ$ west of north at a speed of 4 .8 m/s

Let $V_S$ be the velocity of the ship relative to the earth and $V_{sg}$ be the velocity of the ship with respect to the Gulf stream

Let the west direction be negative x axis and north direction be positive x axis

Now

$V_{sg}=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\V_s-V_g=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\-4.8\sin5^\circ \vec i+4.8\cos5^\circ \vec j-V_g=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\Vg=-1.68\vec\ i+1.156\vec\ j\ \rm m/s\\$

magnitude of the velocity is given by

$V_g=\sqrt{(1.68^2+1.156^2)}\\V_g=2.03\ \rm m/s\\$

Let the velocity of gulf stream makes an angle $\theta$ with the positive y axis we have

$\tan\theta=\dfrac{1.156}{1.68}\\\theta=34.5^\circ$

Velocity of Gulf Stream $34.5^\circ$ west of north at a speed of $2.03\ \rm m/s$

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2 years ago

An experiment is conducted in which red light is diffracted through a single slit. Listed below are alterations made, one at a t

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Answer:

B. The distance between the slits and the screen is halved.

C. The slit width is doubled.

D. A green, rather than red, light source is used.

E. The experiment is conducted in a water-filled tank.

Explanation:

As we know that the position of first minimum is given as

$a sin\theta = N\lambda$

so we have

$\theta = sin^{-1}(\fracN\lambda}{a})$

so width of minimum is given as

$w = L\times sin^{-1}(\fracN\lambda}{a})$

now if we need to decrease the angular position of minimum

1). so we can decrease the distance of screen from the slit

2). we can decrease the wavelength

3). We can increase the width of the slit

So correct answer will be

B. The distance between the slits and the screen is halved.

C. The slit width is doubled.

D. A green, rather than red, light source is used.

E. The experiment is conducted in a water-filled tank.

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2 years ago

Assume that you stay on the earth's surface. what is the ratio of the sun's gravitational force on you to the earth's gravitatio

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First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg.

F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N

Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m

Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356 N

Ratio = 0.356 N/589.18 N
<em>Ratio = 6.04</em>

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2 years ago

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What is the magnitude of the force a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away?

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The magnitude of the force<span> a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away is 1920 Newtons. The formula used to solve this problem is:

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where:
F = Electric force, Newtons
k = Coulomb's constant, 9x10^9 Nm^2/C^2
q1 = point charge 1, C
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Using direct substitution, the force F is determined to be 1920 Newtons.</span>

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2 years ago

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The

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Answer:

(a) Steel rod: $1.1 * 10^{-4}$

Copper rod: $1.88 * 10^{-4}$

(b) Steel rod: $8.3 * 10^{-5} m$

Copper rod: $1.41 * 10^{-4} m$

Explanation:

Length of each rod = 0.75 m

Diameter of each rod = 1.50 cm = 0.015 m

Tensile force exerted = 4000 N

(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as

Strain = $\frac{1}{Y} * \frac{F}{A}$

where Y = Young modulus

F = Fore applied

A = Cross sectional area

For the steel rod:

Y = 200 000 000 000 $N/m^{2}$

F = 4000N

A = $\pi r^{2}$ (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

∴ $Strain = \frac{4000}{200000000000 * 0.000177} \\\\Strain = \frac{4000}{35400000}\\ \\Strain = 0.000113 = 1.13 * 10^{-4}$

For the copper rod:

Y = 120 000 000 000 N/m²

F = 4000N

A = $\pi r^{2}$ (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

$Strain = \frac{4000}{120 000 000 000 * 0.000177} \\\\Strain = \frac{4000}{21240000}\\ \\Strain = = 1.88 * 10^{-4}$

(b) We can find the elongation by multiplying the Strain by the original length of the rods:

Elongation = Strain * Length

For the steel rod:

Elongation = $1.1 * 10^{-4} * 0.75 = 8.3 * 10^{-5} m$

For the copper rod:

Elongation = $1.88 * 10^{-4} * 0.75 = 1.41 * 10^{-4} m$

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2 years ago