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3 years ago

9

a typical top fuel dragstar has a mass of 900 kg what is the force applied to a dragstar by the static friction of the strip the

pushes it down road? f = ma

1 answer:

AnnZ [28]3 years ago

6 0

Wonderful.
You've given us ' m '.
Now, if we only had ' a ', we could come back at you with ' f '.

You might be interested in

If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular

erastova [34]

Answer;

D. The car would begin to move in the direction it was headed in a straight line.

Explanation;

-Centripetal force is any net force causing uniform circular motion. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration.

-The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

-Therefore,If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced then the car would begin to move in the direction it was headed in a straight line.

6 0

3 years ago

Read 2 more answers

Which statement is true for a car that first goes around a curve of radius r at a constant speed v, and then goes around the sam

sasho [114]

Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around

3 0

3 years ago

A car is traveling at 10 m/s. 10 seconds later the car is traveling 40 m/s. What is the car’s acceleration?

ratelena [41]

Answer:

a = 3 m/s^2

Explanation:

Vi = 10 m/s

Vf = 40 m/s

t = 10 s

Plug those values into the following equation:

Vf = Vi + at

40 = 10 + 10a

---> a = 3 m/s^2

3 0

3 years ago

When you eat a candy bar, your body oxidizes glucose to get energy. where does this energy come from? kinetic energy of the gluc

babymother [125]

The correct answer is
<span>"chemical bonds within the glucose molecules "

The chemical bonds of the glucose molecules contain chemical energy, and when these bonds are broken by the processes acting inside the body, the energy of the bonds is released and it can be used by the body.

</span>

8 0

3 years ago

A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten

babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

$P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2$

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0

3 years ago