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Mnenie [13.5K]
3 years ago
10

A 30 g bullet moving a horizontal velocity of 500 m/s comes to a stop 12 cm within a solid wall. (a) what is the change in the b

ullet's mechanical energy? (b) what is the magnitude of the average force from the wall stopping it?
Physics
1 answer:
Sidana [21]3 years ago
7 0
M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet

By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
      = 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.

The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J

If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN

Answer:
(a) 3750 J
(b) 62.5 kN

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15) On a cold day, you take in 4.2 L (i.e., 4.2 x 10-3 m3) of air into your lungs at a temperature of 0°C. If you hold your brea
Rudiy27

Answer:

4.8L ( i.e 4.8 x 10^-3 m3)

Explanation:

Step 1:

Data obtained from the question.

Initial volume (V1) = 4.2L

Initial temperature (T1) = 0°C

Final temperature (T2) = 37°C

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below

K = °C + 273

T1 = 0°C = 0°C + 273 = 273K

T2 = 37°C = 37°C + 273 = 310K

Step 3:

Determination of the final volume.

Since the pressure is constant,

Charles' Law equation will be applied as shown below:

V1 /T1 = V2/T2

4.2/273 = V2 /310

Cross multiply to express in linear form

273 x V2 = 4.2 x 310

Divide both side by 273

V2 = (4.2 x 310)/273

V2 = 4.8L ( i.e 4.8 x 10^-3 m3)

Therefore, the volume of the air in the lungs at that point is 4.8L ( i.e 4.8 x 10^-3 m3)

3 0
3 years ago
What is the net power needed to change the speed of a 1600-kg sport utility vehicle from 15.0 m/s to 40.0 m/s in 4.00 seconds
Sergio039 [100]

Answer:

The net power needed to change the speed of the vehicle is 275,000 W

Explanation:

Given;

mass of the sport vehicle, m = 1600 kg

initial velocity of the vehicle, u = 15 m/s

final velocity of the vehicle, v = 40 m/s

time of motion, t = 4 s

The force needed to change the speed of the sport vehicle;

F = \frac{m(v-u)}{t} \\\\F = \frac{1600(40-15)}{4} \\\\F = 10,000 \ N

The net power needed to change the speed of the vehicle is calculated as;

P_{net} = \frac{1}{2} F[u + v]\\\\P_{net} = \frac{1}{2} \times 10,000[15 + 40]\\\\P_{net} = 275,000 \ W

3 0
2 years ago
Two streams merge to form a river. One stream has a width of 8.4 m, depth of 3.5 m, and current speed of 2.2 m/s. The other stre
hoa [83]

Answer:

Explanation:

We shall solve this problem on the basis of pinciple that water is incompressible so volume of flow will be equal at every point .

rate of volume flow of one stream

= cross sectional area x velocity

= 8.4 x 3.5 x 2.2 = 64.68 m³ /s

rate of volume flow of other stream

= 6.6 x 3.6 x 2.7

= 64.15 m³ /s

rate of volume flow of rive , if d be its depth

= 11.2 x d x 2.8

= 31.36 d

volume flow of river = Total of volume flow rate of two streams

31.36 d  = 64.15 + 64.68

31.36 d  = 128.83

d = 4.10 m /s .

6 0
2 years ago
A proton is placed in an electric field of intensity 700 N/C. What are the magnitude and direction of the acceleration of this p
olya-2409 [2.1K]

Answer:

Acceleration of proton will be a=0.67\times 10^{11}m/sec^2

Explanation:

We have given a proton is placed in an electric field of intensity of 700 N/C

So electric field E = 700 N/C

Mass of proton m=1.67\times 10^{-27}kg

Charge on proton e=1.6\times 10^{-19}C

So electric force on the proton F=qE=1.6\times 10^{-19}\times 700=1.120\times 10^{-16}N

This force will be equal to force due to acceleration of the proton

According to newton's law force is given by F = ma

So 1.67\times 10^{-27}\times a=1.120\times 10^{-16}

a=0.67\times 10^{11}m/sec^2

So acceleration of proton will be a=0.67\times 10^{11}m/sec^2

6 0
3 years ago
I need help with #25 ASAP .. I have to get it done today.. I need help with #25 right now... I'm not playing no games right now
Flura [38]

Answer: 8 or 9

Explanation: they are so many ocean water in the world

6 0
2 years ago
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