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Ksenya-84 [330]
11 months ago
9

a 0.60 kg ball is moving horizontally with a speed of 5.7 m/s when it strikes a vertical wall. the ball rebounds with a speed of

2.7 m/s. what is the magnitude of the change in linear momentum of the ball?
Physics
1 answer:
elixir [45]11 months ago
5 0

We can use the following equation to solve this problem:

<h3>Δp=m∣Vi−Vf∣</h3>

(Where p is the momentum change, m is mass, Vi is initial velocity, and

Vf is final velocity)

Substitute the values given to us in the question:

Δp=(0.6)∣(5.7−(−2.7)∣

Δp=4.9 kg.m/s

Final answer: 4.9kg.m/s.

The change in momentum (Δp) is defined as the change in the mass times the velocity of an object. A force is required to change the momentum of an object. This applied force can increase or decrease momentum, or even change the orientation of an object.

Learn more about momentum:

https://brainly.in/question/1535814

#SPJ4

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Mickey walks Pluto 4 miles North to the dog park. Mickey finds beautiful
yanalaym [24]
Answer is 12 milesBecause yeah
3 0
3 years ago
If a 5000-kg is moving at a speed of 43 m/s, what is its momentum?
Eduardwww [97]

Answer:

215000kgm/s

Explanation:

Given parameters:

Mass of the moving body  = 5000kg

Velocity  = 43m/s

Unknown:

Momentum  = ?

Solution:

The momentum of a body is the amount of motion a body possess.

 It is mathematically expressed as:

  Momentum  = mass x velocity

 Now:

  Momentum  = 5000 x 43  = 215000kgm/s

4 0
3 years ago
A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height
rewona [7]

Answer:

W=17085KJ

Explanation:

From the question we are told that:

Height H=16m

Radius R=3

Height of water H_w=9m

Gravity g=9.8m/s

Density of water \rho=1000kg/m^3

Generally the equation for Volume of water is mathematically given by

 dv=\pi*r^2dy

 dv=\frac{\piR^2}{H^2}(H-y)^2dy

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 dw=(pdv)g (H-y)

Substituting dv

 dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)

 dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy

Therefore

 W=\int dw

 W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy

 W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)

 W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0

 W=3420.84*0.25[2401-65536]

 W=17084965.5J

 W=17085KJ

 

'

'

4 0
3 years ago
A car has a mass of 1000 kg and accelerates at 2 m/s2. What net force is exerted on the car?
prisoha [69]

Answer:

1000N

Explanation:

Based on force=mass*acceleration, if the acceleration is constant at 2 metres per second squared, 1,000kg*2m/s^2=2,000N of force.

If the acceleration steadily increases to 2m/s^2 in 20 seconds, take the average which is 1m/s^2 therefore force=1,000N

6 0
3 years ago
A student starts a food fight by throwing a 0.5 kg burrito at some girl he likes. He throws it kind-of hard so he accelerates it
Elenna [48]

Explanation:

m = mass of burrito thrown by the student = 0.5 kg

a = acceleration of the burrito thrown by the student = 3 m/s²

F = force applied by the student on the burrito = ?

According to newton's second law , the net force on an object is the product of its mass and acceleration. it is given as

F = ma

inserting the values

F = (0.5) (3)

F = 1.5 N

hence the net force on the burrito comes out to be 1.5 N

4 0
3 years ago
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