All categories

3 years ago

In a ballistics test, a 52g bullet hits a sand bag and stops after moving 1.34 m. If the initial bullat

Physics

1 answer:

olya-2409 [2.1K]3 years ago

8 0

Answer:

Friction force on the bullet is 58.7 N opposite to its velocity

Explanation:

As we know that initial speed of the bullet is 55 m/s

after travelling into the sand bag by distance d = 1.34 m it comes to rest

so final speed

$v_f = 0$

now we can use kinematics top find the acceleration of the bullet

$v_f^2 - v_i^2 = 2 a d$

so we have

$0 - 55^2 = 2(a)(1.34)$

$a = -1128.7 m/s^2$

now by Newton's II law we know that

$F = ma$

so we have

$F = (0.052)(-1128.7)$

$F = -58.7 N$

You might be interested in

An air bubble has a volume of 2.0 cm3 when it is released by a submarine 100 m below the surface of a freshwater lake. What is t

UkoKoshka [18]

Answer:

21.35 cm^3

Explanation:

let the volume at the surface of fresh water is V.

The volume at a depth of 100 m is V' = 2 cm^3

temperature remains constant.

density of water, d = 1000 kg/m^3

Pressure at the surface of fresh water is atmospheric pressure,

P = Po = 1.013 x 10^5 N/m^2

The pressure at depth 100 m is P' = Po + hdg

P' = $1.013 \times 10^{5}+ 100 \times 1000 \times 9.8$

P' = 10.813 x 10^5 N/m^2

Use the Boyle's law

P V = P' V'

$1.013 \times 10^{5}\times V = 10.813 \times 10^{5}\times 2$

V = 21.35 cm^3

Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.

5 0

3 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

4 years ago

One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e

frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is $1.122 X 10^{-7} kg$

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

$E= \Delta mc^{2}$

where $\Delta m$ = change in mass

c = speed of light = $3 \times 10 ^{8}m/s$

Making m the subject of the formula, we can find the change in mass to be

$\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg$

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

5 0

3 years ago

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses $x_1$ distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1$

When speed doubles

$\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2$

divide 1 and 2

$\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}$

$x_2=2x_1$

Therefore spring compresses twice the initial value

7 0

3 years ago

18. How does magnetic force affect the energy in a system of magnets when the magnets are released and begin to move? a The magn

astra-53 [7]

When the magnets are released and begin to move, the magnetic force changes potential energy into kinetic energy;<u> option D</u>

<h3>What is magnetic force?</h3>

Magnetic force can be defined as the force of attraction or repulsion that is felt or produced between the magnetic poles and electrically charged moving particles.

Magnetic force is the force that exists due to the attractio or repulsion of objects place around the region of space where a maget exerts its force.

When a magnet is released ai a system of magnets and allowed to move it increases the kinetic eergy of the system by converting potential energy to kinetic energy.

Learn more about magnetic force at: brainly.com/question/28989998

#SPJ1

5 0

1 year ago