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Inessa [10]
2 years ago
12

A 1000-kg sports car of mass accelerates from rest to 20 m/s in 6.6 s. What is the frictional force exerted by the road on the c

ar? Group of answer choices 1500 N 1750 N 2750 N 3000 N
Physics
1 answer:
lukranit [14]2 years ago
4 0

The frictional force exerted by the road on the car is 3000N

Given the mass of the car is 1000 kg , the velocity of the car is 20m/s

and the time is 6.6s

We need to find the frictional force exerted by the road on the car

We know that Force = mass * acceleration

Now here Mass is given but acceleration is not given

So, we will find acceleration by using the formula v = u+at

Where u = 0

v = 20m/s

a = ?

t = 6.6

Substitute the values in the formula We get

v = u+at

20 = 0+(a)(6.6)

20/6.6 = a

∴ a = 3.03 m/s^2

Rounding to nearest tenth is 3m/s^2

Hence the acceleration is 3m/s^2

Now substituting the value of acceleration in F = ma

Where m = mass

a = acceleration

F = 1000*3

∴ F = 3000N

Hence the frictional force exerted by the road on the car is 3000 N

Learn more about Frictional force here

brainly.com/question/23161460

#SPJ4

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When you look at the backside of a shiny teaspoon held at arm's length, do you see yourself upright or upside down? (b) When you
igor_vitrenko [27]

Answer:

a) The back spoon gives a right image (upright)

b) the front gives an inverted image

Explanation:

The spoon is a curved metallic object, when we see ourselves from the back we have a convex mirror, in this type of mirror when the law of reflection is applied the rays diverge therefore the eye-brain system forms the image with the prolongation of the rays, therefore the image is straight and smaller than the object.

When we look through the deep side of the spoon, we have a concave mirror and as the object (we) is further away than the distance, the rays converge to a point, so the image is real, inverted smaller than the object.

In summary.

a) The back spoon gives a right image (upright)

b) the front gives an inverted image

3 0
2 years ago
Which of the following are not vector directions?
Svetradugi [14.3K]

Answer:

here north are not vector option b hope ur help

6 0
3 years ago
Read 2 more answers
A object with a mass of 1.5 kg is lifted from the ground to a height of 0.22 m what is the objects potential energy
svet-max [94.6K]

Answer:

<h2>3.3 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 1.5 × 10 × 0.22

We have the final answer as

<h3>3.3 J</h3>

Hope this helps you

5 0
2 years ago
A 1.15 kg book is at rest on the table. What is the magnitude of the normal force that the table is exerting on the book?
Agata [3.3K]

Answer:

11.27N

Explanation:

Given parameters:

Mass of the book  = 1.15kg

Unknown:

Magnitude of the normal force  = ?

Solution:

The normal force is the vertical force exerted by a body on an object.

It can be described as the weight of an object.

 Normal force  = mass x acceleration due to gravity

 Normal force  = 1.15 x 9.8  = 11.27N

5 0
2 years ago
The tongue weight of a trailer should be what percent of the gross trailer weight rating
mario62 [17]

Answer:

between 10 and 15 percent

Explanation:

How to put your load

- First load the heavy

The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.

In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.

-  Hold your load

After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.

- Trailer weight

To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.

GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.

7 0
3 years ago
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