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3 years ago

For your answer to this problem, just type in the numerical magnitude of the momentum - no units.

Physics

1 answer:

stepan [7]3 years ago

3 0

Answer:

120 kg•m/s.

Explanation:

From the question given above, the following data were obtained:

Case 1

Mass of object = M

Velocity of object = V

Momentum = 15 kg•m/s

Case 2

Mass of object = 2M

Velocity of object = 4V

Momentum = ?

Momentum is defined as follow:

Momentum = mass × velocity

The momentum of object in case 2 can be obtained as follow:

From case 1

Momentum = mass × velocity

15 = M × V

15 = MV ....... (1)

From case 2:

Momentum = mass × velocity

Momentum = 2M × 4V

Momentum = 8MV ....... (2)

Finally , substitute the value of MV in equation 1 into equation 2.

Momentum = 8MV

MV = 15

Momentum = 8 × 15

Momentum = 120 kg•m/s

Therefore, an object with a mass of 2M and 4V would have a momentum of 120 kg•m/s

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A student examines a 20-meter long rectangular stream channel and takes the following measurements: width of stream = 4 meters,

Scrat [10]

Answer:

The discharge of the stream at this location is 40 cubic meters per second.

Explanation:

The discharge is the volume flow rate of the water in the stream. For this purpose we can use the following formula:

Discharge = Volume Flow Rate = (Cross-Sectional Area)(Velocity of Stream)

Volume Flow Rate = (Width of Stream)(Depth of Stream)(Velocity of Stream)

Volume Flow Rate = (4 meters)(2 meters)(5 meters per second)

<u>Volume Flow Rate = 40 cubic meters per second</u>

Therefore, the discharge of the stream at this location is found to be <u>40 cubic meters per second</u>

This result shows that 40 cubic meters volume of water passes or discharges through this point in a time of one second. Hence, this is called the volume flow rate or the discharge of the stream.

3 0

3 years ago

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

2 years ago

A tennis racket hits a tennis ball with a force of F=at−bt2, where a = 1290 N/ms , b = 330 N/ms2 , and t is the time (in millise

denpristay [2]

Answer:

The resulting velocity of the ball after it hits the racket was of V= 51.6 m/s

Explanation:

m= 55.6 g = 0.0556 kg

t= 2.8 ms = 2.8 * 10⁻³ s

F= 1290 N/ms * t - 330 N/ms² * t²

F= 1024.8 N

F*t= m * V

V= F*t/m

V= 51.6 m/s

6 0

3 years ago

A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring

sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

4 0

3 years ago

The atoms of a molecule come from two or more?

AlladinOne [14]

Compounds. two or more compounds.

7 0

3 years ago