Ask question

Ask question

All categories

2 years ago

13

Calculate the horizontal force that must be applied to a 1,300 kg vehicle to give it an acceleration of 2.6 m/s² on a level road

.
500 N

3,380 N

1,302.6 N

0.002 N

1 answer:

ANEK [815]2 years ago

4 0

The answer is 3,380 N

You might be interested in

What physical property of light changes and causes the light to refract, or bend?

Tcecarenko [31]

I think it would be by the change in speed
For example if light travels from air to water it will slow down causing it to travel at a different angle/direction

3 0

3 years ago

WHICH IS THE HARDEST SUBSTANCE IN EARTH?

PIT_PIT [208]

Diamond is the hardest naturally occurring substance found on Earth.

8 0

2 years ago

Read 2 more answers

What point does a basketball have the greatest potential energy

Semenov [28]

When it’s about to be dropped

4 0

3 years ago

Read 2 more answers

Technician A says that torque-to-yield head bolts are special bolts that are reusable. Technician A says that main bearing clear

Blizzard [7]

Answer: D

Neither A nor B

Explanation:

In order to check the clearances for rod and main bearings, you need a set of micrometers and a dial-bore gauge

Measuring the inside diameter of a main or rod bearing will require a dial bore gauge. The best ones to use are accurate down to 0.0001-inch.

So, both technician A and B are incorrect

7 0

2 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

5 0

3 years ago