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3 years ago

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A 1.2 kg block of wood hangs motionless from strings. A 50 g bullet, traveling horizontally, strikes the block and becomes embed

ded inside the block. Immediately after the bullet becomes embedded in the block, the block is observed to have a speed of 8.0 m/s. What was the speed of the bullet before it hit the block?
A. 200 m/s.
B. 12 m/s.
C 98 m/s.
D 9,604 m/s.
E 57 m/s.

1 answer:

kicyunya [14]3 years ago

4 0

Answer:

A. <u>200m/s</u>

Explanation:

Using the law of conservation of momentum expressed as;

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses of the object

u1 and u2 are the respective velocities

v is the common velocity

Given

m1 = 1.2kg

u1 = 0m/s (block is a stationary object)

m2 = 50g= 0.05kg

u2 = ?

v = 8.0m/s

Substitute the values into the formula and get u2 (speed of the bullet before hitting the block)

1.2(0)+0.05u2 = (1.2 + 0.05)(8)

0.05u2 = 1.25(8)

0.05u2 = 10

u2 = 10/0.05

u2 = 200m/s

Hence the speed of the bullet before it hit the block is <u>200m/s</u>

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3 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

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Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

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k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

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$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

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3 years ago

Cara is swimming in a river. She can swim downstream up to a maximum speed of 66 meters per-minute. If she is swimming downstrea

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3 years ago

Read 2 more answers

8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott

nasty-shy [4]

<u>Answer:</u> The ball is travelling with a speed of 5.5 m/s after hitting the <u>bottle.</u>

<u>Explanation:</u>

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

$m_1,u_1\text{ and }v_1$ are the mass, initial velocity and final velocity of ball.

$m_2,u_2\text{ and }v_2$ are the mass, initial velocity and final velocity of bottle.

We are given:

$m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s$

Putting values in above equation, we get:

$(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s$

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

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3 years ago