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3 years ago

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A 5.00 g bullet traveling 355 m/s is stopped by lodging in the side of a building. The heat produced is shared between the build

ing and the bullet. How much heat in Joules must be shared?

1 answer:

steposvetlana [31]3 years ago

7 0

Answer:

Explanation:

Given

mass of bullet $m=5\ gm$

speed of bullet $v=355\ m/s$

bullet is stopped by building and heat produced is shared between building and bullet

Kinetic Energy of bullet is converted into Thermal energy

Kinetic Energy of bullet $K=\frac{1}{2}mv^2$

$K=0.5\times 5\times 10^{-3}\times (355)^2$

$K=315.06\ J$

So 315.06 J of Energy is converted in to thermal energy

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Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y

Gnesinka [82]

The representation of this problem is shown in Figure 1. So our goal is to find the vector $\overrightarrow{R}$ . From the figure we know that:

$\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}$

From geometry, we know that:

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

Then using vector decomposition into components:

$For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85$

Therefore:

$R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m$

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector $\overrightarrow{R}$ , that is:
</span>
$\left | \overrightarrow{R} \right |=
\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}$

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span> $\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}$

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3 years ago

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 15.1 K.

Zarrin [17]

Answer: (a) The magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .

(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$ .

Explanation:

(a) Expression for change in temperature is as follows.

$|\Delta T| = |x - y|K$

= 15.1 K

= $|(x - 273.15) - (y - 273.15)|^{o}C$

= $|x - y|^{o}C$

= $15.1^{o}C$

Therefore, the magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .

(b) Change in temperature from Celsius to Fahrenheit is as follows.

F = 1.8C + 32

C = $\frac{F - 32}{1.8}$

Since, K = C + 273

or, $\Delta K = \Delta C = \frac{\Delta F}{1.8}$

$\Delta F = 1.8 \Delta K$

= 1.8 (15.1)

= $27.18^{o}F$

or, = $27.2^{o}F$

Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$ .

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3 years ago

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If your brother changes the channel, you will change it back, so then he will change it, then you will change it, etc, etc. This

KonstantinChe [14]

The answer is the law of inetia

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3 years ago

Calculate the total resistance for a 650ohm , a 350 ohm , and a 1000 ohm resistor connected in series

Mekhanik [1.2K]

Answer:

2000 ohms

Explanation:

Resisters in series just add.

Rt = R1 + R2 + R3

R1 = 650 ohm

R2 = 350 ohm

R3 = 1000 ohm

Rt = 650 + 350 + 1000

Rt = 2000 ohms.

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3 years ago

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A contestants spin a wheel when it is their turn in a game show. One contestant gives the wheel an initial angular speed of 3.40

guajiro [1.7K]

Answer:

4.62 s

Explanation:

We are given that

Initial angular speed, $\omega=3.4 rad/s$

$\theta=1\frac{1}{4} rev=\frac{5}{4}\times 2\pi=2.5\pi rad$

$\omega'=0$

$\omega'^2-\omega^2=2\alpha \theta$

Substitute the values

$0-(3.4)^2=2\times 2.5\pi \alpha$

$\alpha=\frac{-(3.4)^2}{2\times 2.5\pi}=-0.736 rad/s^2$

$\omega'=\omega+\alpha t$

$0=3.4-0.736 t$

$-0.736t=-3.4$

$t=\frac{-3.4}{-0.736}=4.62 s$

Hence, the wheel takes 4.62 s to come to rest.

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3 years ago