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3 years ago

An object traveling at a 5.0 m/s East has a mass of 4.0kg. What is the magnitude of the object’s momentum?

Physics

1 answer:

Shtirlitz [24]3 years ago

6 0

Answer: p = 20 kg.m/s Eastward

Explanation: Momentum is the product of mass and velocity of the object or p= mv.

p = mv

= 4.0 kg ( 5.0 m/s)

= 20 kg.m/s Eastward

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Car A (mass 1100 kg) is stopped at a traffic light when it is rearended by car B(mass 1400 kg). Both cars then slide with locke

viva [34]

Answer:

Part a)

$v_a = 3.94 m/s$

Part b)

$v_b = 3.35 m/s$

Part C)

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

$a = -\frac{F_f}{m}$

$a = -\frac{\mu mg}{m}$

$a = - \mu g$

now we will have

$v_f^2 - v_i^2 = 2ad$

$0 - v_i^2 = 2(-\mu g)(6.1)$

$v_a = \sqrt{(2(0.13)(9.81)(6.1)}$

$v_a = 3.94 m/s$

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

$v_b = \sqrt{2(0.13)(9.81)(4.4)}$

$v_b = 3.35 m/s$

Part C)

By momentum conservation we will have

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1400 v_b = 1100(3.94) + 1400(3.35)$

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0

3 years ago

Gold has a density of 19300 kg/m3 calculate the mass of 0.02m3 of gold in kilograms

aev [14]

Answer:

The mass of 0.02 m³ of gold is 386 kilograms

Explanation:

Given:

The density of the gold = 19300 kg/m³.

The volume of gold = 0.02 m³

To Find:

The mass of gold = ?

Solution:

We know that density is mass divided per unit volume.

Thus mathematically

Density = \frac{mass}{volume}Density=

volume

mass

Rewriting in terms of mass ,

Mass = density * volume

On substituting the known values

Mass = 19300 kg/m³ * 0.02 m³

Mass = 386 kilograms

Learn more about Mass and Density:

Mass=?,volume=190,density=4

Mass 350 kg volume 175 density ans

This is not my answer I copied it but hope it helps:)

5 0

3 years ago

What unit of measure would you use for mass?

creativ13 [48]

grams, pounds, kilos, etc

6 0

3 years ago

A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the

Shalnov [3]

Heat required to raise the temperature of a given system is

$Q = ms\Delta T$

here we know that

m = mass

s = specific heat capacity

$\Delta T$ = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

$Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2$

$Q = 5 * 1700 * 25 + 2 * 900 * 25$

$Q = 212500 + 45000$

$Q = 257500 J$

So heat required to raise the temperature of the system is 257500 J

4 0

3 years ago

PLEASE ANSWER QUICKLY

Rom4ik [11]

Answer:

a luminous ball of plasma

6 0

2 years ago