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2 years ago

Take the mechanical equivalent of heat as 4 J/cal. A 10-g bullet moving at 2000 m/s plunges

1 answer:

4 0

Assuming that all the bullet’s energy heats the paraffin, its final temperature is 27.1 degree C. The correct option is D.

<h3>What is temperature?</h3>

Temperature is the degree of hotness and coldness of the material.

The energy of the bullet E = 1/2 mv²

E = 1/2 x 10 x 10⁻³ x (2000)²

E = 2 x 10⁴ J

This heat is used in heating the paraffin

E = m x c ΔT = m x c (Tfinal -Tinitial)

2 x 10⁴ J = 1 x 2.8 x 10³ x (Tfinal -20)

Tfinal = 27.1°C

Thus, the final temperature is 27.1 degree C. The correct option is D.

Learn more about temperature.

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What is the name of the particle that compounds are made of?

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Explanation:

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3 years ago

Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

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Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$

The sum of the vertical components of F1 and F2 must equal the total force Ft

$\displaystyle -2F\ cos21^o+F\ sin\alpha =460$

Solving for $\alpha$ in the first equation

$\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o$

$\displaystyle cos\alpha =0.717=>\alpha =44.214^o$

$\displaystyle F(2cos21^o+sin\alpha)=460$

$\displaystyle F=\frac{460}{2cos21^o+sin\alpha}$

$\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}$

$\displaystyle F=179.37\ N$

The magnitude of F1 is

$|F1|=2*F=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

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3 years ago

What is another name for the breathing mechanism?

aleksandrvk [35]

Aspirate or inhale or respire or

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3 years ago

A certain car battery with a 12.0 V emf has an initial charge of 131 A · h. Assuming that the potential across the terminals sta

irga5000 [103]

Answer:

The battery can supply 130 W for 11.75 h

Explanation:

In order to discover the time in wich the battery can supply this energy we need to find how much current is being drawn from it, we do that by using the equation for real power that is P = V*I, since we have V and P we can solve for I as seen bellow:

I = P/V = 130/12 = 10.834 A

We can use this value to find how many hours the power can supply said current. We do that by dividing the current capacity of the battery by the current drawn:

t = 141/12 = 11.75 h

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3 years ago

A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A

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Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

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2 years ago