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3 years ago

A 2-kg ball is thrown at 3 m/s. What is the ball's momentum? *

Physics

1 answer:

crimeas [40]3 years ago

8 0

Answer:

Given

mass (m) =2kg

velocity (v) =3m/s

momentum (p) =?

Form

p=mv

2kgx3m/s

p=6kg.m/s

the momentum of ball's =6kg.m/s

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What kind of charge is produced when a silk shirt is rubbed with human body<br>

jeka94

Answer:

Frictional charges

Explanation:

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3 years ago

A 2.0 kg block is pulled across a horizontal surface by a 15 N force at a constant velocity. What is the force of friction actin

lianna [129]

Answer:

<em>The force of friction acting on the block has a magnitude of 15 N and acts opposite to the applied force.</em>

Explanation:

<u>Net Force </u>

The Second Newton's law states that an object acquires acceleration when an unbalanced net force is applied to it.

The acceleration is proportional to the net force and inversely proportional to the mass of the object.

If the object has zero net force, it won't get accelerated and its velocity will remain constant.

The m=2 kg block is being pulled across a horizontal surface by a force of F=15 N and we are told the block moves at a constant velocity. This means the acceleration is zero and therefore the net force is also zero.

Since there is an external force applied to the box, it must have been balanced by the force of friction, thus the force of friction has the same magnitude acting opposite to the applied force.

The force of friction acting on the block has a magnitude of 15 N opposite to the applied force.

6 0

2 years ago

The graph shows the progress of a car which traveled 300 km in 5 hrs. Assuming that its speed remained constant, how far had the

lisov135 [29]

If the car travels
300 km.........................5 hrs
?km ..............................2h
to find out we do (300*2)/5=600/5=120km
also
300km/5h=60km/h
in 2 hours
120km

6 0

3 years ago

Example 3 :

kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d = diameter = 120mm = 0. 12m

V = velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × $\frac{0. 06}{0. 12* 1* 0. 9}$

friction factor = 0. 52 × $\frac{0. 06}{0. 108}$

friction factor = 0. 52 × 0. 55

friction factor $= 0. 289$

b. When V = 3mls

Friction factor = 0. 52 × $\frac{0. 06}{0. 12 * 3* 0. 9}$

Friction factor = 0. 52 × $\frac{0. 06}{0. 324}$

Friction factor = 0. 52 × 0. 185

Friction factor $= 0.096$

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × $\frac{1}{2*6. 6743 * 10^-11}$ × $\frac{1^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 1. 80 × 10^8

Head loss When V = 3m/s

Head loss = $0. 096$ × $\frac{1}{1. 334 *10^-10}$ × $\frac{3^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

#SPJ1

4 0

1 year ago

12. AABC is a right triangle. If AB = 3 and AC = 7, find BC. Leave your answer in simplest radical form.

Dennis_Churaev [7]

Answer: A 2 square root 3

Explanation:

4 0

3 years ago