1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kogti [31]
3 years ago
5

Why do cars stay in motion.

Physics
2 answers:
Oksi-84 [34.3K]3 years ago
7 0

Answer:

it stays in motion

Explanation:

An object in motion will stay in motion until a force acts on it

liubo4ka [24]3 years ago
6 0

Answer:

Friction is the resistance to motion created by two objects moving against each other. Friction creates heat. Unless acted on by a force, objects in motion tend to stay in motion and objects at rest remain at rest.

You might be interested in
A tennis ball is served horizontally from 2.4m above the ground at net is 12m away and point 0.9 high will be ball clear the net
Schach [20]

Explanation:

Let us first calculate  long does it take to go 12m at 30m/s( assumed speed)

12/30 = 0.4 seconds

horizontal distance the ball drop in that time

H= (0)(0.4)+1/2(-9.8)(0.4)2

H= -0.78m

negative sign shows that the height of the ball at the net from the top.

Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m

As 1.62m>0.9m so the ball will clear the net.

H_1= V0y t’ + ½ g t’^2

-2.4= (0)t’ + ½ (-9.8) t’^2

t’= 0.69s

X’=V0x t’

X’=(30)(0.96)

X’= 20.7m

3 0
3 years ago
Two charged objects, A and B, are exerting an electric force on each other. What will happen if the charge on A is increased?
ozzi

Answer: The forces acting on both of them will increase in magnitude.

Explanation:

According to Coulomb's law, the electrostatic force between two bodies is proportional to the product of their two charges. If the charge on A is increased this product increases in size (it must have been non-zero to begin with, since there was a force between them at first). Thus, the force between them rises.

6 0
3 years ago
You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce
Aleksandr [31]

The time needed for the 7th car to pass is 13.2 s

Explanation:

The motion of the train is a uniformly accelerated motion, therefore we can use suvat equations.

We start by analzying the motion of the first car, by using the equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered by the first car in a time t, which corresponds to the length of one car

u = 0 is the initial velocity

a is the acceleration

t = 5.0 s is the time

The equation can be rewritten as

a=\frac{2s}{t^2}=\frac{2L}{(5.0)^2}=0.08L[m/s^2]

where L is the length of one car.

The same equation can be written considering the first 7 cars:

7L = ut+\frac{1}{2}at'^2

where

7L is the distance covered by the 7 cars

t' is the time needed

We still have

u = 0

And the acceleration is constant so it is

a=0.08L

Substituting into the equation, we can find t':

7L = \frac{1}{2}(0.08L)t'^2\\7=0.04t'^2\\t'=\sqrt{\frac{7}{0.04}}=13.2 s

In attachment the graph of the distance covered versus the time taken: since the motion is uniformly accelerated, the relationship between the two variables is quadratical.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
How far will it go, given that the coefficient of kinetic friction is 0.10 and the push imparts an initial speed of 3.9 m/s ?
Akimi4 [234]

Answer:19.5 m

Explanation:

Given

coefficient of kinetic Friction \mu _k=0.10

Initial speed u=3.9\ m/s

Friction is present so it tries to stop to the object and stops it completely after moving certain distance let say s

maximum deceleration provided by friction is

a_{max}=\mu _k\cdot g

a_{max}=0.1\times 3.9=0.39\ m/s^2

using equation of motion

v^2-u^2=2 as

where v=final\ velocity

u=initial velocity

a=acceleration

s=displacement

(0)-(3.9)^2=2\times (-0.39)\times s

s=19.5\ m

6 0
3 years ago
Which of these is a possible disadvantage of using wind as a source of
gizmo_the_mogwai [7]

Answer:

D. Wind turbines take up a lot of space.

Explanation:

In wind turbines the kinetic energy received by the air molecules is converted into electrical energy by the use of turbines

So here in order to get more kinetic energy from air we need more crossectional area of the wind mill to interact with the air

So here we need the large size of turbines

so this is the main disadvantage of the wind turbines because it needs large area to install the whole setup also the efficiency of this turbine is small so it needs large number of wind mills to setup good output power

so correct answer will be

D. Wind turbines take up a lot of space.

3 0
3 years ago
Read 2 more answers
Other questions:
  • Which occurrence would lead you to conclude that lights are connected in a
    13·1 answer
  • What is current measured in​
    8·1 answer
  • a kid jumping on a trampoline reaches a height of 0.925m. what was his speed when he left the trampoline?
    10·1 answer
  • A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru
    12·1 answer
  • Variations in the reactive properties of different organic molecules are most closely associated with _____.
    13·2 answers
  • In a reaction the mass of a reactant decreases by 58g in 233 seconds. What is the rate?
    6·1 answer
  • Justin strikes a 0.058-kg golf ball with a force of 260 N. If the ball moves with a velocity of 63 m/s, calculate the time the b
    6·1 answer
  • 12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it
    7·1 answer
  • HELP ASAP!! WILL TRY TO GIVE BRAINLIEST
    15·1 answer
  • Think about the way optical fibers are designed. why are the ends of the fibers glowing very brightly while the sides of the fib
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!