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A bullet of mass 11.1 g is fired into an initially stationary block and comes to rest in the block. The block, of mass 1.01 kg,

Kryger [21]

Answer:

a) The initial speed of the bullet is 488 m/s

b) The loss of kinetic energy is 1.3 × 10³ J.

Explanation:

Hi there!

To solve this problem we have to use the conservation of momentum:

initial momentum of the bullet + initial momentum of the block =

final momentum of the block-bullet system

The momentum of an object is calculated as follows:

p = m · v

Where:

p = momentum

m = mass of the object.

v = velocity.

Then, in our system:

p₁₁ = initial momentum of the bullet.

p₂₁ = initial momentum of the block.

p₃₂ = final momentum of the block-bullet system.

p₁₁ + p₂₁ = p₃₂

The initial momentum of the bullet will be:

p₁₁ = m · v

p₁₁ = 0.0111 kg · v

The initial momentum of the block will be:

p₂₁ = 1.01 kg · 0 m/s = 0 kg · m/s

The final momentum of the block-bullet system will be:

p₃₂ = (1.01 kg + 0.0111 kg) · 5.30 m/s

Then, by conservation of the momentum:

initial momentum of the bullet = momentum of the block-bullet system

0.0111 kg · v = (1.01 kg + 0.0111 kg) · 5.30 m/s

v = ((1.01 kg + 0.0111 kg) · 5.30 m/s)/ 0.0111 kg

v = 488 m/s

The initial speed of the bullet is 488 m/s

b) The initial kinetic energy (KE) of the system is the kinetic energy of the bullet because the block is at rest:

KE = 1/2 · m · v²

KE = 1/2 · 0.0111 kg · (488 m/s)²

KE = 1.32 × 10³ J

The final kinetic energy of the system will be the kinetic energy of the block-bullet system:

KE = 1/2 · (1.01 kg + 0.0111 kg) · (5.30 m/s)²

KE = 14.3 J

The loss of kinetic energy will be:

initial kinetic energy - final kinetic energy

1.32 × 10³ J - 14.3 J = 1.3 × 10³ J

The loss of kinetic energy is 1.3 × 10³ J.

8 0

3 years ago

A waist-to-hip ratio of 1.0 is considered healthy.<br><br> True<br> False

laila [671]

false a waist to hip ratio that is healthy would be .80 or less is considered healthy for most individuals hope this helps

7 0

3 years ago

Read 2 more answers

Phase, color, and ductility are all examples of what type of property?

Mama L [17]

Physical properties.

7 0

3 years ago

Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem

statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

$x=x_o+v_1t$ (1)

xo = 10 km

v1 = 55km/h

car 2:

$x'=v_2t$ (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

$x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}$

$t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s$

The position in which both cars coincides is:

$x=(55km/h)(0.33h)=18.33km$

6 0

3 years ago

Two men are standing on a frictionless ice surface holding opposite ends of a rope. One man (mass = 80 kg) pulls on the rope wit

ankoles [38]

Answer:

The acceleration of man 1 and 2 is $3.125\ m/s^2$ and $4.167\ m/s^2$ .

Explanation:

Mass of man 1, m₁ = 80 kg

Mass of man 2, m₂ = 60 kg

One man pulls on the rope with a force of 250 N.

Let a₁ is acceleration of man 1,

F = m₁a₁

$a_1=\dfrac{F}{m_1}\\\\a_1=\dfrac{250}{80}\\\\a_1=3.125\ m/s^2$

Let a₂ is acceleration of man 1,

F = m₂a₂

$a_2=\dfrac{F}{m_2}\\\\a_2=\dfrac{250}{60}\\\\a_2=4.167\ m/s^2$

So, the acceleration of man 1 and 2 is $3.125\ m/s^2$ and $4.167\ m/s^2$ .

7 0

3 years ago