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krek1111 [17]
3 years ago
11

1-A car with momentum 19016 kg*m/s has a mass of 1300kg. What is the speed

Physics
1 answer:
Ymorist [56]3 years ago
6 0
The speed would be 130 mph because the car is very heavy
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Pleaseee HElpp!!!!!!!
lutik1710 [3]

Coulomb's Law

Given:

F = 3.0 x 10^-3 Newton

d = 6.0 x 10^2 meters

Q1 = 3.3x 10^-8 Coulombs

k = 9.0 x 10^9 Newton*m^2/Coulombs^2

Required:

Q2 =?

Formula:

F = k • Q1 • Q2 / d²

Solution:

So, to solve for Q2

 

Q2 = F • d²/ k • Q1

Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9 Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)

Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)

Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)

Then, take the reciprocal of the denominator and start multiplying

Q2 = 1080 • 1 Coulombs/297

Q2 = 1080 Coulombs / 297

Q2 = 3.63636363636 Coulombs

Q2 = 3.64 Coulumbs

6 0
3 years ago
Read 2 more answers
What is the value of work done on an object when a 70 newton force moves it 9.0 meters in the same direction as the force
olchik [2.2K]
<span>Work, very simply, equals force times distance (when the force and distance are in the same direction. otherwise you get a little bit of trig added on) \[W=F*\Delta x\] W=70N * 9.0 m = 630 Nm = 630 J</span>
3 0
2 years ago
Read 2 more answers
What is the moment of inertia of a disc of mass 5kg and radius 10cm?
vodka [1.7K]

Answer:

500

Explanation:

6 0
3 years ago
Read 2 more answers
Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
Phoenix [80]

Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

7 0
3 years ago
The two waves shown below pass through the same medium in the phases shown and interfere with each other.
ale4655 [162]

A. W

Explanation:

The wave that would be produced by the interaction of the two waves shown in the diagram is wave W.

There is no wave in the diagram W.

This type of interference is known as destructive interference.

  • Destructive interference occurs when two waves out of phase comes together.
  • In this way, they cancel out each other and are terminated.
  • If the two waves are in phase, they will reinforce one another.
  • When waves reinforce one another, a constructive interference has occurred.

learn more:

Color in soap bubbles brainly.com/question/8733443

#learnwithBrainly

5 0
3 years ago
Read 2 more answers
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