This is all of my points

How much force is needed to stop a 80 kg football player if he decelerates at 5m/s?

cricket20 [7]

Use Newton's 2nd law of motion.
Here it is:

   Force = (mass) x (acceleration)

   = (80 kg) x (-5 m/s²)

   =    -400 Newtons .

I called the acceleration negative because the player is slowing down.

The force comes out negative because it's in the direction opposite to
his motion.

4 0

4 years ago

A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?

dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0

3 years ago

Distance between bholu and golu house house 9 km . bholu has to attend golu birthday party at 7am . he started from his home at

navik [9.2K]

For the rest 3 km

Bholu need to run in 5 minutes so

He need to 3 × 1000 / 5 × 60 m/s

3000/ 300 = 10 m/s

So , his speed should be 1 m/s to reach in 5 minute

mark as brainliest 5 star

6 0

4 years ago

Read 2 more answers

Consider a motor that exerts a constant torque of 25.0 N⋅m to a horizontal platform whose moment of inertia is 50.0 kg⋅m2 . Assu

Step2247 [10]

To solve this exercise it is necessary to apply the concepts related to Work and Kinetic Energy. Work from the rotational movement is described as

$W=\tau \Delta\theta$

In the case of rotational kinetic energy we know that

$KE = \frac{1}{2}I\omega^2$

PART A) $\theta$ is given in revolutions and needs to be in radians therefore

$\theta = 12rev(\frac{2\pi rad}{1rev})$

$\theta = 24\pi rad$

Replacing in the work equation we have to

$W=\tau \Delta\theta$

$W= (25)(24\pi)$

$W = 1884.95J$

PART B) From the torque and moment of inertia it is possible to calculate the angular acceleration and the final speed, with which the kinetic energy can be determined.

$\tau = I \alpha$