All categories

2 years ago

A blue puck has a velocity of 3i –4j m/s. Its mass is 20 kg. What is its momentum?

Physics

2 answers:

damaskus [11]2 years ago

6 0

P = m * v
v = {3i - 4j} = square root (3^2 + 4^2) = 5
P = 20 * 5
P = 100 kg m/s

Tom [10]2 years ago

3 0

Magnitude:

√(3² + 4²) = √(9 + 16) = √25 = 5

Momentum = (mass) x (velocity) = (20kg x 5m/s) = 100 kg-m/s

Direction:

3i - 4j is in the direction tan⁻¹(-4/3) = 53.1 south of east.

You might be interested in

The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

A freight train has a mass of $1.83\times 10^{7}\,kg$ . The wheels of the locomotive push back on the tracks with a constant net force of $7.50\times 10^{5}\,N$ , so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?

If locomotive have a constant net force ( $F$ ), measured in newtons, then acceleration ( $a$ ), measured in meters per square second, must be constant and can be found by the following expression:

$a = \frac{F}{m}$ (1)

Where $m$ is the mass of the freight train, measured in kilograms.

If we know that $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , then the acceleration experimented by the train is:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now, the time taken to accelerate the freight train from rest ( $t$ ), measured in seconds, is determined by the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - Final speed of the train, measured in meters per second.

$v_{o}$ - Initial speed of the train, measured in meters per second.

If we know that $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ and $v = 22.222\,\frac{m}{s}$ , the time taken by the freight train is:

$t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }$

$t = 542.265\,s$

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0

1 year ago

You stand on a bathroom scale in a moving elevator. what happens to the scale reading if the cable holding the elevator suddenly

Viefleur [7K]

A bathroom scales works due to gravity. Under normal conditions, a reading can be obtained when your body is pushing some force on the scale. However in this case, since you and the scale are both moving downwards, so your body is no longer pushing on the scale. Therefore the answer is:

The reading will drop to 0 instantly

7 0

2 years ago

Read 2 more answers

1. Each year at a college, there is a tradition of having a hoop rolling competition. Alex rolls his 0.350 kg hoop down the cour

grigory [225]

Question 1:

Answer:

The moment of inertia of Alex's rolling hoop is 0.197 $kg \cdot cm^2$

Explanation:

Given:

Mass of the hoop = 0.350 g

Radius of the hoop = 75.0 cm

To Find:

The moment of inertia of Alex's rolling hoop = ?

Solution:

The moment of inertia = $mr^2$

where

m is the mass

r is the radius

Converting cm to m, we get

75.0 cm = 0.75 m

Now substituting the values,

=> moment of inertia = $(0.350)(0.75)^2$

=> moment of inertia = $(0.350)(0.5625)$

=> moment of inertia = $(0.197)$

Question 2:

Answer:

The combined angular momentum of the masses is 1.76 $kg m^2 s^{-1}$

If she pulls her arms in to 0.12 m, her new linear speed is $18.33 m/s^2$

Explanation:

Given:

Mass = 2.0 kg

Radius = 0.8 m

Velocity = 1.2 m/s

a.The combined angular momentum of the masses:

$L = r \cdot m \cdot v_1$

Substituting the values,

$L = 0.8 \cdot 2.0 \cdot 1.1$

L= 1.76 $kg m^2 s^{-1}$

b. If she pulls her arms in to 0.12 m, what is her new linear speed

$0.12 \cdot 0.8 \cdot v_2 = 1.76$

$0.096 cdot v_2 = 1.76$

$v_2 = \frac{1.76}{0.096}$

$v_2 = 18.33 m/s^2$

6 0

1 year ago

Given that the CFL bulb is rated at 10% efficiency, if the LED bulb consumes half the amount of electrical power, for the same a

Butoxors [25]

Answer and explanation:

The efficiency of the LED bulb is 20% while The efficiency of the Inc bulb is 1%.

The lightbulb efficiency can be defined as the quotient between the amount of usable light and the amount of electrical power output. For the efficiency of the CFL Bulb:

$\epsilon_{CFL}=\frac{L_0}{P_0}=0.1$

Therefore:

$\epsilon_{led}=\frac{L_0}{0.5P_0}=2 \cdot0.1=0.2\\\epsilon_{INC}=\frac{L_0}{10P_0}=0.1 \cdot 0.1=0.01$

4 0

1 year ago

Arm abcd is pinned at b and undergoes reciprocating motion such that θ=(0.3 sin 4t) rad, where t is measured in seconds and the

storchak [24]

θ=0.3sin(4t)
w=0.3cost(4t)(4)=1.2cost(4t)
a=-4.8sin(4t)

cos4t max will always be 1 (refer to cos graph), for same reason, sin4t will always be 0

therefore, wmax=1.2rad/s

vAmax=r*w=250*1.2=300mm/s
(may be different if your picture/radius is from a different picture)

adt=a*r=200*-4.8sin(4t)=0 (sin(4t)=0)

adn=r*w^2=200*1.2^2=288

ad= square root of adt^2+adn^2 = 288mm/s^2

8 0

1 year ago