a train traveling at 30m/s comes to a complete stop in 20 sec.how far did the train go during the deceleration

The particle is an electron. the field slows down the electron without deflecting it. what is the direction of the electric fiel

sammy [17]

The particle is an electron. The field slows down the electron without deflecting it. The direction of the electric field is <u>right.</u>

In physics, the motion of electrically charged particles gives rise to a field called an electric field. It is measured in force per unit charge.

This field applies force on other charged particles.

Particles bearing opposite charges attract each other while particles having similar charges repel each other in the field.

If a positive charge is placed in the field then the field line moves in an outward direction and for a negative charge, the direction of the lines is inward.

If you need to learn more about electric field click here:

brainly.com/question/26199225

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7 0

2 years ago

A speed-time graph is shown below:

Juliette [100K]

Answer:

It traveled 4 centimeters.

Explanation:

In a speed versus time graph, the distance travelled is given by the area under the graph.

In this graph we have the following:

- The speed of the object is v = 1 cm/s between time t = 0 s and t = 4 s

- The speed of the object is v = 0 cm/s between time t = 4 s and t = 8 s

Since the speed in the second part is zero, the distance travelled in the second part is zero. So, the only distance travelled by the object is the distance travelled during the first part, which is equal to the area of the first rectangle:

$d=v\Delta t=(1)(4-0)=4 cm$

4 0

3 years ago

If the distance to a distant galaxy is recorded as 10,000 light years, then light must have been traveling for 5000 years to rea

satela [25.4K]

False false false false

5 0

2 years ago

Read 2 more answers

ASAP! AWARDING BRAINLIEST see attached photo.

stellarik [79]

Answer:d

Explanation:

6 0

3 years ago

Read 2 more answers

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

5 0

3 years ago