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Hunter-Best [27]

2 years ago

12

The momentum of an object is determined to be 7.2 × 10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (

Note: 1 kg = 1000 g).

1 answer:

slavikrds [6]2 years ago

6 0

Answer:

Momentum, p = 7.2 g-m/s

Explanation:

It is given that,

The momentum of an object is $p=7.2\times 10^{-3}\ kg-m/s$

We need to express momentum in any equivalent units. There can be many solutions of this problem. Some of the units of mass are gram, milligram etc. units of length are meters, mm etc.

Since, 1 kg = 1000 gram

So, $p=7.2\times 10^{-3}\times 10^3\ g-m/s$

Therefore, the momentum of the object is 7.2 g-m/s. Hence, this is the required solution.

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If the intensity level by 15 identical engines in a garage is 100 dB, what is the intensity level generated by each one of these

insens350 [35]

To develop this problem it is necessary to apply the concepts related to Sound Intensity.

By definition the intensity is given by the equation

$\beta = 10Log(\frac{I}{I_0})$

Where,

I = Intensity of Sound

$I_0$ = Intensity of Reference

At this case we have that 15 engines produces 15 times the reference intensity, that is

$I= 15I_0$

And the total mutual intensity is 100 dB, so we should

$\beta = 100-10*log(\frac{15I_0}{I_0})$

$\beta = 100-10*log(15)$

$\beta = 100-11.76$

$\beta = 88.23dB$

Therefore each one of these engines produce D. 88dB.

5 0

2 years ago

A force pair is produced when a tennis racket strikes a tennis ball. Which of the following best explains why the tennis ball do

Serga [27]

Answer:

Each half of the force pair acts on a different object.

Explanation:

When a tennis racket strikes a tennis ball a pair force is produced. when the racket strikes the ball the racket exerts an action force on the tennis ball, according to Newton's third law for every action there is an equal and opposite reaction force, as a reaction the ball exert an equal and opposite force on the racket. These forces are often called pair forces.

As the forces acts on different bodies (Action force act on ball and reaction force act on racket) so the net force tennis ball is never zero.

4 0

2 years ago

If a young protostar with a disk is rotating and shrinking. how much faster is it rotating after its size has decreased by a fac

maks197457 [2]

In this system we have the conservation of angular momentum: L₁ = L₂
We can write L = m·r²·ω

Therefore, we will have:
m₁ · r₁² · ω₁ = m₂ · r₂² · ω₂

The mass stays constant, therefore it cancels out, and we can solve for ω<span>₂:
</span>ω₂ = (r₁/ r₂)² · ω<span>₁

Since we know that r</span>₁ = 4r<span>₂, we get:
</span>ω₂ = (4)² · ω<span>₁
= 16 </span>· ω<span>₁

Hence, the protostar will be rotating 16 </span><span>times faster.</span>

5 0

2 years ago

Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o

IRINA_888 [86]

Answer:

a) $W=-0.0103125\ J$

b) $W=0.0059375\ J$

c) Compressing is easier

Explanation:

Given:

Expression of force:

$F=kx-bx^2+cx^3$

where:

$k=100\ N.m^{-1}$

$b=700\ N.m^{-2}$

$c=12000\ N.m^{-3}$

$x$ when the spring is stretched

$x$ when the spring is compressed

hence,

$F=100x-700x^2+12000x^3$

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, $x=-0.05\ m$

Now,

The spring constant at this instant:

$j=\frac{F}{x}$

$j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}$

$j=-8.25\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times -8.25\times (-0.05)^2$

$W=-0.0103125\ J$

b)

When compressing the spring by 0.05 m

we have, $x=0.05\ m$

<u>The spring constant at this instant:</u>

$j=\frac{F}{x}$

$j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}$

$j=4.75\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times 4.75\times (0.05)^2$

$W=0.0059375\ J$

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0

1 year ago

A test car carrying a crash test dummy accelerates from 0 to 30 m/s and then crashes into a brick wall. Describe the direction o

earnstyle [38]

Answer:

the direction of acceleration of the vehicle is the same direction of its velocity of car

s acceleration has the opposite direction to the car speed.

Explanation:

The initial acceleration of the car can be calculated with

v = v₀ + a t

a = (v-v₀) t

indicate that the initial velocity is zero (v₀ = 0 m / s)

a = v / t

a = 300 / t

the direction of acceleration of the vehicle is the same direction of its acceleration movement.

When the car collides with the wall, it exerts a force in the opposite direction that stops the vehicle, therefore this acceleration has the opposite direction to the car speed. But your module must be much larger since the distance traveled to stop is small

5 0

2 years ago