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xxTIMURxx [149]
2 years ago
8

A 5.4 g lead bullet moving at 261 m/s strikes a steel plate and stops. If all its kinetic energy is converted to thermal energy

and none leaves the bullet, what is its temperature change? Assume the specific heat of lead is 128 J/kg · ◦ C. Answer in units of ◦C.
Physics
1 answer:
scZoUnD [109]2 years ago
8 0

Answer:

Change in temperature ∆(tita) is 266.097°C

Explanation:

Ok kinectic energy = 1/2MV²

5.4 grams =( 5.4/1000) kilogram

Kinectic energy =( 1/2 )*(5.4/1000)*261²

Kinectic energy = 183.9267 joules

If kinetic energy = thermal energy

183.9267 joules = mc∆(tita)

Where ∆(tita) = change in temperature

And c = 128 J/kg

∆(tita) = 183.9267/((5.4/1000)*128)

∆(tita) = 266.097

∆(tita) = 266.097°C

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Compared to its weight on Earth, a 5kg object on the moon will weigh
shutvik [7]

Answer:

8.1 N/49 N=0.1653  which means 16.53% of the weight of the object on Earth.

Explanation:

On the Moon, where the gravitational constant is 1.62 \frac{m}{s^2}, the weight of the 5 kg object will be: weight_M=m*g_M = 5 kg * 1.62 \frac{m}{s^2} =8.1 N

Where the answer is in Newtons (N) since all quantities are given in the SI system.

On Earth, on the other hand, the weight of the object is:

weight_E=m*g_E= 5 kg* 9.8 \frac{m}{s^2} = 49N

Therefore the object's weight on the Moon compared to that on Earth will be:

8.1N/49N=0.1653

That is, 16.53% of the weight the object has on Earth.

5 0
3 years ago
An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Exp
Artemon [7]

Explanation:

It is given that,

Mass of the football player, m = 92 kg

Velocity of player, v = 5 m/s

Time taken, t = 10 s

(1) We need to find the original kinetic energy of the player. It is given by :

k=\dfrac{1}{2}mv^2

k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2

k = 1150  J

In two significant figure, k=1.2\times 10^3\ J

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

i.e. P=\dfrac{W}{t}=\dfrac{\Delta K}{t}

P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}

P = 115 watts

In two significant figures, P=1.2\times 10^2\ Watts

Hence, this is the required solution.  

6 0
2 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the
Rama09 [41]

1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

And now we can find the energy stored,which is given by:

U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

and using

E=4\cdot 10^5 V/m, we find

u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

7 0
3 years ago
Which is an example of an exothermic process?
Scorpion4ik [409]

Answer:

an example of an exthermic process is combustion

Explanation:

combustion is like lighting a candle

3 0
3 years ago
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Answer:

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8 0
3 years ago
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