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NemiM [27]
4 years ago
7

In a physics experiment, two equal-mass carts roll towards each other on a level, low-friction track. One cart rolls rightward a

t 2 m/s and the other cart rolls leftward at 1 m/s. After the carts collide, they couple (attach together) and roll together with a speed of __________. Ignore resistive forces.a. 0.5 m/s
b. 0.33 m/s
c. 0.67 m/s
d. 1.0 m/s
e. none of these
Physics
1 answer:
xz_007 [3.2K]4 years ago
7 0

Answer:

The correct answer is option a.

Explanation:

Conservation of momentum :

m_1u_1+m_2u_2=m_1v_1+m_1v_2

Where :

m_1, m_2 = masses of object collided

u_1,u_2 = initial velocity before collision

v_1,v_2 = final velocity after collision

We have :

Two equal-mass carts roll towards each other.

m_1=m_2=M

Initial velocity of m_1=u_1=2 m/s

Initial velocity of m_2=u_2=-1 m/s (opposite direction)

Final velocity of m_1=v_1=v (same direction )

Final velocity of m_2=v_2=v  (same direction)

M\times 2 m/s+M(-1 m/s)=Mv+Mv

1 m/s=2v

v = 0.5 m/s

rg135

The speed of the carts after their collision is 0.5 m/s.

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The frequency of the second harmonic is 880 Hz (a pitch of A5). The speed of sound through the pipe is 350 m/sec. Find the frequency of the first harmonic and the length of the pipe. The length of an air column is related mathematically to the wavelength of the wave which resonates within it.

4 0
3 years ago
Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the
Over [174]

Answer:

Part a)

a= 0.32 m/s^2

Part b)

F_c = 3.6 N

Part c)

F_c = 5.5 N

Explanation:

Part a)

As we know that the friction force on two boxes is given as

F_f = \mu m_a g + \mu m_b g

F_f = 0.02(10.6 + 7)9.81

F_f = 3.45 N

Now we know by Newton's II law

F_{net} = ma

so we have

F_p - F_f = (m_a + m_b) a

9.1 - 3.45 = (10.6 + 7) a

a = \frac{5.65}{17.6}

a= 0.32 m/s^2

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

F_c - F_f = m_b a

F_c = \mu m_b g + m_b a

F_c = 0.02(7)(9.8) + 7(0.32)

F_c = 3.6 N

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

F_p - F_f - F_c = m_b a

9.1 - 0.02(7)(9.8) - F_c = 7(0.32)

F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)

F_c = 5.5 N

3 0
3 years ago
A 1232 kg car moving north at 25.6 m/s collides with a 2028 kg car moving north at 17.5 m/s . They stick together. In what direc
Citrus2011 [14]

Answer:

I. Angle = 41.7° Northeast.

II. Vr = 7.08m/s

Explanation:

Let the two cars be denoted by A and B

<u>Given the following data;</u>

Mass of car A = 1232 Kg

Velocity of car A = 25.6 m/s

Mass of car B = 2028 Kg

Velocity of car B = 17.5m/s

First of all, we would solve for momentum;

Momentum = mass × velocity

Momentum, M1 = 1232 × 25.6

Momentum, M1 = 31539.2 Kgm/s

Momentum, M2 = 2028 × 17.5

Momentum, M2 = 35490 Kgm/s

Now, let's find the resultant momentum using the Pythagoras theorem;

R² = M1² + M2²

R² = 31539.2² + 35490²

R² = 994721136.6 + 1259540100

R² = 2254261237

Taking the square root of both sides, we have

Resultant momentum, R = 47479.06 Kgm/s

To find the direction;

Angle = tan¯¹(M1/M2)

Angle = tan¯¹(31539.2/35490)

Angle = tan¯¹(0.89)

<em>Angle = 41.7° Northeast.</em>

To find the speed;

R = (M1 + M2)Vr

47479.06 = (31539.2 + 35490)Vr

47479.06 = 67029.2Vr

Vr = 47479.06/67029.2

<em>Vr = 7.08m/s</em>

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Answer:

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Answer:

over harvesting, pollution, introduction of invasive species such as new predators and food competitors

Explanation:

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The five mass extinction events are

Ordovician Mass Extinction

Devonian Mass Extinction

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