All categories

4 years ago

A 5000 kg truck traveling at 60 m/s stops in 5 seconds. How much friction was between the truck's tires and the ground?

Physics

1 answer:

oksano4ka [1.4K]4 years ago

8 0

Answer:

60000N

Explanation:

acceleration is change in velocity

a =(v-u)/t where a is acceleration u is initial velocity and v is final velocity

a = (0-60)/5 = -60/5= - 12m/s^2

here minus sign shows that body is decelerating and force is force of friction Now f = ma here f is force of friction m is mass and a is acceleration

f= 5000×- 12= -60000N

MINUS SIGN HERE SHOWS FORCE OF FRICTION

Hence force of friction is 60000N

You might be interested in

A scientific hypothesis can be disproved with

11111nata11111 [884]

It can be disproved with <span>one reproducible experiment.</span>

3 0

3 years ago

Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.3×109 V and the quantity of charg

Georgia [21]

Answer:

A) ΔU = 3.9 × 10^(10) J

B) v = 8420.75 m/s

Explanation:

We are given;

Potential Difference; V = 1.3 × 10^(9) V

Charge; Q = 30 C

A) Formula for change in energy of transferred charge is given as;

ΔU = QV

Plugging in the relevant values gives;

ΔU = 30 × 1.3 × 10^(9)

ΔU = 3.9 × 10^(10) J

B) We are told that this energy gotten above is used to accelerate a 1100 kg car from rest.

This means that the initial potential energy will be equal to the final kinetic energy since all the potential energy will be converted to kinetic energy.

Thus;

P.E = K.E

ΔU = ½mv²

Where v is final velocity.

Plugging in the relevant values;

3.9 × 10^(10) = ½ × 1100 × v²

v² = [7.8 × 10^(8)]/11

v² = 70909090.9090909

v = √70909090.9090909

v = 8420.75 m/s

4 0

3 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago

You're sitting in the back of a taxi and place your bag on the seat next to you. The taxi enters a traffic circle and travels th

svp [43]

Answer:

(d) III only

Explanation:

We have to observe the motion of the bag with respect to taxi , considering taxi as stationary or inertial frame . Since bag is not moving with respect to taxi , the inertial frame that means , net force on it is zero .So option i and ii are ruled out .

Now how to explain motion of the bag ie why it is stationary ie what are the balancing force acting on it. We know that on a body on circular path , a force called centripetal force is acting on it . So that force must be acting on it . The balancing force is the frictional force which is keeping it stationary with respect to taxi . Hence the third option is correct.

5 0

3 years ago

_________ is stored energy.

Ivan

Answer:

I think so is b kinetic o a potential

5 0

3 years ago

A 5000 kg truck traveling at 60 m/s stops in 5 seconds. How much friction was between the truck's tires and the ground? ​

A 5000 kg truck traveling at 60 m/s stops in 5 seconds. How much friction was between the truck's tires and the ground?