All categories

3 years ago

A cart is uniformly decelerating from rest. The net force acting on the cart is

Physics

1 answer:

mina [271]3 years ago

7 0

Answer:

i think the answer is constant

You might be interested in

A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i v, st

OLEGan [10]

Answer:

The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Explanation:

The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.

When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.

So the centripetal acceleration, a' = 16v²/R.

To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16

a'/a = 16

a' = 16a.

So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

8 0

4 years ago

PLEASE SOMEONE HELP ME ILL GIVE THE PERSON BRAINLIEST

Yuliya22 [10]

I believe it would be c

5 0

3 years ago

A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. What is the initial momentum of the g

soldi70 [24.7K]

Answer:

I think answer is zero

bcz momentum=mass×velocity

body was initially at rest it means its velocity is zero

30×0=0

6 0

3 years ago

A car runs for about 10 years during its average useful life. If the odometer reads 120,000 miles at the end, what was the avera

Eva8 [605]

Answer:

12000 mph

Explanation:

Given that,

Distance read by the odometer = 120,000 miles

Duration of car, t = 10 years

We need to find the average speed of the car. Speed of an object is equal to the total distance covered divided by total time taken. So,

$v=\dfrac{120000}{10}\\\\=12000\ mph$

So, the average speed of the car is 12000 mph.

7 0

3 years ago

Can someone help me?

Leviafan [203]

Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

$V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t$

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

$V_y.t = \frac{2d}{t} .t = 2d$

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

4 0

3 years ago