Answer this question i will mark you brainliest

Copper has a specific heat of 387 J/kg°C. A copper bullet is moving at 950 m/s. If the bullet is suddenly stopped so that all of

Alla [95]

Answer:

1196.02 °C

Explanation:

If the kinetic energy is converted into heat,

then,

Kinetic energy of the copper = heat energy of the copper

1/2m(v²) = cm(t₂-t₁)

where m = mass of copper, v = velocity of copper, c = specific heat capacity of copper, t₂ = final temperature of copper, t₁ = initial temperature of copper.

Since the mass of copper remains the same,

1/2v² = c(t₂-t₁)

make t₂ the subject of the equation

t₂ = 1/2(v²/c)+t₁..................... Equation 1

Given: v = 950 m/s, c = 387 J/kg°C, t₁ 30 °C

Substitute into equation 1

t₂ = 1/2(950²/387)+30

t₂ = 1196.02 °C

Hence the temperature the bullet reach before it was stopped = 1196.02 °C

8 0

4 years ago

In your own words, explain how Doppler radar works. Describe the properties of electromagnetic waves and interactions that make

lisov135 [29]

If you stand up in a big room and echo, your voice will echo from the walls. As long as the room is empty. Since the speed of sound is constant, depending on air density, the more humid the air the faster and farther sound travels. The speed of sound is constant, you could measure the time it takes for your voice to echo off the walls. The same thing happens with Doppler radar, but it’s not voice, it has higher frequency signals.<span> </span>

6 0

3 years ago

Read 2 more answers

At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magne

larisa [96]

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

$F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j$

F = ma

∴

$a ^{\to}= \dfrac{F^{\to}}{m}$

$a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)$

$a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j$

At time t(sec; the partiCle velocity becomes $v(t) = 3.78 v_o$

The velocity of the charge after the time t(sec) is expressed by using the formula:

$v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}$

8 0

3 years ago

2. A 200.0-kg bear grasping a vertical tree slides down at constant velocity. What is the friction force between the tree and th

Neporo4naja [7]

Answer: 1960 N

Explanation:

The bear is sliding down at constant velocity: this means that its acceleration is zero, so the net force is also zero, according to Newton's second law:

$F_{net}=ma=0$

There are two forces acting on the bear: its weight W, pulling downward, and the frictional force Ff, pulling upward. Therefore, the net force is given by the difference between the two forces:

$F_{net}=W-F_f=0$

From the previous equation, we find that the frictional force is equal to the weight of the bear:

$F_f=W=mg=(200.0 kg)(9.8 m/s^2)=1960 N$

8 0

3 years ago

What is the peak emf generated (in V) by rotating a 1000 turn, 42.0 cm diameter coil in the Earth's 5.00 ✕ 10−5 T magnetic field

ELEN [110]

Answer:

5.77 Volt

Explanation:

N = 1000, Diameter = 42 cm = 0.42 m, t = 12 ms = 12 x 10^-3 s

Change in magnetic field, B = 5 x 10^-5 T

The peak value of emf is given by

e = N x dФ / dt

e = N x A x dB/dt

e = (1000 x 3.14 x 0.21 x 0.21 x 5 x 10^-5) / (1.2 x 10^-3)

e = 5.77 Volt

3 0

3 years ago

Answer this question i will mark you brainliest​

Answer this question i will mark you brainliest