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WITCHER [35]
3 years ago
12

Ballistic data obtained on a firing range show that aerodynamic drag reduces the speed of a .44 magnum revolver bullet from 250

m/s to 210 m/s as it travels over a horizontal distance of 150 m. The diameter and mass of the bullet are 11.2 mm and 15.6 g, respectively. Evaluate the average drag coefficient for the bullet.
Physics
1 answer:
m_a_m_a [10]3 years ago
4 0

Answer:

0.363999909622

Explanation:

F = Force

m = Mass = 15.6 g

C = Drag coefficient

ρ = Density of air = 1.21 kg/m³

A = Surface area = \dfrac{\pi}{4}d^2

v = Terminal velocity = v=210\ m/s

s = Displacement = 150 m

a=\dfrac{v^2-u^2}{2s}

Force is given by

F = ma

F=\dfrac{1}{2}\rho CAv^2\\\Rightarrow ma=\dfrac{1}{2}\rho CAv^2\\\Rightarrow m\dfrac{v^2-u^2}{2s}=\dfrac{1}{2}\rho CAv^2\\\Rightarrow C=2\times m\dfrac{v^2-u^2}{2s}\times\dfrac{1}{\rho Av^2}\\\Rightarrow C=2\times15.6\times 10^{-3}\dfrac{210^2-250^2}{2\times 150}\times\dfrac{1}{1.21\times\dfrac{\pi}{4}\times (11.2\times 10^{-3})^2(210)^2}\\\Rightarrow C=-0.363999909622

The drag coefficient is 0.363999909622 (ignoring negative sign)

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A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force on the wire wh
zheka24 [161]

Answer:

Magnetic force, F = 0.24 N

Explanation:

It is given that,

Current flowing in the wire, I = 4 A

Length of the wire, L = 20 cm = 0.2 m

Magnetic field, B = 0.6 T

Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :

F=ILB\ sin\theta

F=4\ A\times 0.2\ m\times 0.6\ T\ sin(30)

F = 0.24 N

So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.

7 0
4 years ago
Read 2 more answers
If the sprinter accelerates at that rate for a distance of 15 m, and then maintains the velocity he has at that point for the re
ahrayia [7]

Answer:

The time for the entire race is 11.39 sec.

Explanation:

Given that,

Distance = 15 m

Remainder distance = 100 m

Suppose A sprinter begins a race with an acceleration of 3.4 m/s².

We need to calculate the time

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value in the equation

15=0+\dfrac{1}{2}\times3.4\times t^2

t^2=\dfrac{30}{3.4}

t=2.97\ sec

We need to calculate the final velocity of sprinter

Using equation of motion again

v=u+at

Put the value into the formula

v=0+3.4\times2.97

v=10.09\ m/s

We need to calculate the distance covers by sprinter

d=100-15=85\ m

The sprinter need to covers only 85 m.

We need to calculate the time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t'=\dfrac{85}{10.09}

t'=8.42\ sec

We need to calculate the time for the entire race

t''=t+t'

Put the value into the formula

t''=2.97+8.42

t''=11.39\ sec

Hence, The time for the entire race is 11.39 sec.

4 0
3 years ago
How long will it take to go 150km traveling at 50km/hr? step by step please
Zigmanuir [339]
To find out time, you put distance over speed. So you would have to put 150 over 50. You divide 150 by 50 and you would get 3. So your answer is 3 hours.
3 0
3 years ago
Identify the conditions for an elastic collision in a closed system. Check all that apply.
s344n2d4d5 [400]

Answer:

In an elastic collision:

  • There is no external net force acting. Thus, Momentum before and after collision is equal. Momentum remains conserved.
  • Total energy always remains conserved as energy cannot be created nor destroyed. It can change from one form to another.
  • There is no lost due to friction in elastic collision. So the kinetic energy is also conserved.
  • Velocities may change after collision. If the masses are equal, the velocities interchange.

When one object is stationary:

Final velocity of object 1:

v₁ = (m₁ - m₂)u₁/(m₁ +m₂)

Final velocity of object 2:

v₂ = (2 m₁ u₁)/(m₁+m₂) =

  • Objects do not stick together in elastic collision. They stick together in inelastic collision.
  • One object may be stationary before the elastic collision.

Thus, conditions for an elastic collision:

  • Energy is conserved.
  • Velocities may change.
  • Momentum is conserved.
  • Kinetic energy is conserved.
  • One object may be stationary before the elastic collision.
7 0
3 years ago
Read 2 more answers
If a force of 12 N is applied to an object
Varvara68 [4.7K]

Answer:

<h3>The answer is 3 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{12}{4}  \\

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

6 0
3 years ago
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