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Softa [21]
3 years ago
12

An object is moving at 60m/s and has a mass of 5 kg what is its momentum​

Physics
1 answer:
charle [14.2K]3 years ago
5 0

Answer:

300

Explanation:

the momentum is 300

p=mv

p=5×60

5×60 =300

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A 61 kg skater is traveling at 2.5 m/s while carrying a 4.0 kg bowling ball. After he throws the bowling ball forward at twice t
gregori [183]

The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

p_i = (M+m)u

where

M = 61 kg is the mass of the skater

m = 4.0 kg is the mass of the ball

u = 2.5 m/s (forward) is the combined velocity of the skater and the ball

After, the ball is thrown at twice the velocity, so the final total momentum is

p_f = MV+mv

where

V is the final velocity of the skater

v = 2(2.5) = 5.0 is the final velocity of the ball

Since the total momentum must be conserved, we can write

p_i = p_f\\(M+m)u = MV+mv\\V=\frac{(M+m)u-mv}{M}=\frac{(61+4.0)(2.5)-(4.0)(5.0)}{61}=2.34 m/s

So, the skater is moving at 2.34 m/s (forward) after the shot.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

5 0
4 years ago
A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin
astraxan [27]

Answer:

966 mph

Explanation:

Using as convention:

- East --> positive x-direction

- North --> Positive y-direction

The x- and y- components of the initial velocity of the jet can be written as

v_{1x} = 406 mph\\v_{1y} = 0

While the components of the velocity of the wind are

v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph

So the components of the resultant velocity of the jet are

v_x = v_{1x}+v_{2x}=406+548.6=954.6 mph\\v_y = v_{1y}+v_{2y}=0+147.0=147.0 mph

And the new speed is the magnitude of the resultant velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(954.6)^2+(147.0)^2}=965.8 mph \sim 966 mph

6 0
3 years ago
Its not b...................
vladimir1956 [14]

sir what's the question you have all you wrote is it's not b

4 0
3 years ago
A race car makes one lap around a track of radius 50 m in 9.0 s. What is the average velocity? *
Oksi-84 [34.3K]

Given that,

Radius of track, r = 50 m

time , t = 9 s

velocity, v = ?

Distance covered by car in one lap around a track is equal to the circumference of the track.

C = 2 π r = 2 * 3.14 * 50

C = 314.159 m

Distance covered by car, s = 314.159 m

Velocity = distance/ time

V = 314.159 / 9

V = 34.9 m/s

The average velocity of car is 34.9 m/s.

7 0
3 years ago
With a 900 V source, voltage is divided across 3 series resistors of 300 V, 280 V, and:
telo118 [61]

Answer:

AAAAAAAAaaaaaaaaaaaaaaaaaaaaaaaaaaaa

6 0
3 years ago
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