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Allushta [10]
3 years ago
11

During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of

72.0 kg and its horizontal component of velocity is 6.50 m/s just before the 65.0 kg performer catches it. If the performer is initially motionless on nearly frictionless roller skates, what is his speed immediately after catching the cannon ball
Physics
1 answer:
olasank [31]3 years ago
8 0

Answer:

3.416 m/s

Explanation:

Given that:

mass of cannonball m_A = 72.0 kg

mass of performer m_B = 65.0 kg

The horizontal component of the ball initially \mu_{xA} = 6.50 m/s

the final velocity of the combined system v = ????

By applying the linear momentum of conservation:

m_A \mu_{xA}+m_B \mu_{xB} = (m_A+m_B) v

72.0 \ kg \times 6.50 \ m/s+65.0 \ kg \times 0 = (72.0 \ kg+65.0 \ kg) v

468 kg m/s + 0 = (137 kg)v

v = \dfrac{468\  kg m/s }{137 \ kg}

v = 3.416 m/s

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Answer:

A sled and its rider are moving at a speed of along a horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050. What is the displacement x of the sled?

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2 years ago
A disk has a radius of 30 cm and a mass of 0.3 kg and is turning at 3.0 rev/s. A trickle of sand falls onto the disk at a distan
Pani-rosa [81]

Answer:

The mass of the sand that will fall on the disk to decrease the is 0.3375 kg

Explanation:

Moment before = Moment after

I \omega_i = I \omega_f +mr^2 \omega_f\\\\mr^2 \omega_f = I \omega_i  - I \omega_f \\\\m = \frac{ I \omega_i  - I \omega_f}{r^2 \omega_f }

where;

I is moment of inertia = Mr² = 0.3 x (0.3)² = 0.027 kg.m²

substitute this in the above equation;

m = \frac{ 0.027[3(2 \pi)  - 2(2 \pi)]} {0.2^2 * 6\pi } = \frac{ 0.027[6 \pi  - 4\pi]} {0.2^2 * 4\pi }\\\\m = 0.3375kg

Therefore, the mass of the sand that will fall on the disk to decrease the is 0.3375 kg

7 0
3 years ago
If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is th
Taya2010 [7]

Answer: The magnitude of the current in the second wire 2.67A

Explanation:

Here is the complete question:

Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?

Explanation: Please see the attachments below

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3 years ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

We know that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, temperature on left side is 1.48 times the temperature on right

6 0
3 years ago
Can an ordinary object, like a motorcycle, be mass-less? Yes or No
Drupady [299]

Answer:

no.

Explanation:

because the mass of an object never changes.

4 0
3 years ago
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