All categories

2 years ago

a moving truck takes a much longer time to stop than that taken by a car when brakes are pressed at the same time

Physics

1 answer:

Radda [10]2 years ago

6 0

Answer:

hi here is your answer

Explanation:

a moving truck takes longer time to stop because it has more inertia and the truck has greater Momentum also

----------HOPE IT HELPS----------

You might be interested in

The irreversible thermite reaction (Fe2O3 + 2 Al --> 2 Fe + Al2O3) is highly exothermic, changing the temperature by over 100

likoan [24]

Answer:

the water will boil and change into a gas

Explanation:

This is because it is exothermic and releases the energy into the surroundings and the fact that water boils at 100 degrees Celsius

6 0

3 years ago

Which of the following is not a part of dalton s atomic theory?

Alex777 [14]

<span>c. atoms are always in motion..............</span>

6 0

3 years ago

Read 2 more answers

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

puteri [66]

Answer:

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is $0\ mph\ to\ 70\ mph$ .

Initial velocity of the car, $v_i=0\ mph$

final velocity of the car during the test, $v_f=32\ mph=14.3052\ m.s^{-1}$

Time taken to accelerate form zero to 32 mph at full power, $t=1.2\ s$

initial velocity of the car, $u_i=0\ mph$

final desired velocity of the car, $u_f=64\ mph=28.6105\ m.s^{-1}$

Now the acceleration of the car:

$a=\frac{v_f-v_i}{t}$

$a=\frac{14.3052-0}{1.2}$

$a=11.921\ m.s^{-1}$

Now using the equation of motion:

$u_f=u_i+a.t$

$64=0+11.921\times t$

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0

3 years ago

If you pull up on a bucket with a tension force of 15 N and the bucket has a weight of 15 N, what is the net force acting on the

fredd [130]

Answer:

F = 0 [N]

Explanation:

To solve this problem we must perform a summation of forces in the direction of the vertical axis. Where the positive force is that of the tension of the upward force, while the force exerted by the weight is directed downward with a negative sign.

ΣF = 0

15 - 15 + F = 0

F = 0 [N]

3 0

2 years ago

When a board with a box on it is slowly tilted to a larger and larger angle, common experience shows that the box will at some p

Savatey [412]

Answer:

c. The coefficient of kinetic friction is less than the coefficient of static friction

Explanation:

When the box finally does break loose. Then the component of the box weight which is parallel to the board weight parallel component, is equal to the $\mu_gn$ .

$w_{box}=\mu_gn$

For the box to acce;erate thee must be non-zero net force acting on the box parallel to the board. Or we can say,

$w_1>f_g\\w_1>\mu_gn$

Therefore the force of kinetic friction must be less than the force of static friction. Thus,

$\mu_g$

3 0

2 years ago

a moving truck takes a much longer time to stop than that taken by a car when brakes are pressed at the same time​

a moving truck takes a much longer time to stop than that taken by a car when brakes are pressed at the same time