Ask question

Ask question

All categories

4 years ago

12

If Michael Jordan has a vertical leap of 1.29 meters, what is his take-off speed and his hang time (total time to move upwards t

o the peak and then return to the ground)?

1 answer:

evablogger [386]4 years ago

6 0

(Hint: the time<span> to rise to the </span>peak<span>is one-half the </span>total hang-time<span>.).</span>

You might be interested in

A horizontal force of 20 N is applied to the object. Will the force applied make the object move?

goldenfox [79]

It depends on two factors: The direction of application of force and the coefficient of friction, $\mu$ , between the object and the surface on which the object is placed.

If the direction of application of force is vertically downwards, then the object will not move for any value of $\mu$ .

If the coefficient of friction is zero, $\mu=0$ , then any value of force that is not in a vertically downward direction, the object will move.

If the direction of application of force is not vertically downwards, then it depends on the value of frictional force.

8 0

3 years ago

An elevator filled with passengers has a mass of 1.70×103kg. (a) The elevator accelerates upward from rest at a rate of 1.20m/s2

Anton [14]

Answer with Explanation:

We are given that mass of an elevator filled with passengers = $1.70\times 10^3 Kg$

a.a= $1.2 m/s^2$

t=1.50 s

Initial velocity=0

We have to calculate the tension in the cable supporting the elevator.

According to newton's second law,the net force is given by

$F_net=\sum F=ma$

$F-mg=ma$

Substitute the values then we get

$F=ma+mg=m(a+g)=1.7\times 10^3(1.2+9.8)=1.87\times 10^4 N$

Hence, the tension in the cable supporting the elevator= $1.87\times 10^4 N$

b.t=8.5 s

We have to find the tension in the table during this time.

Fnet=0

F=W= $1.7\times 10^3\times 9.8=1.67\times 10^4 N$

Hence, the tension in the cable during this time = $1.67\times 10^4 N$

c. Deceleration= $0.6 m/s^2$

t=3 s

We have to find the tension in the cable during deceleration.

$F-mg=-ma$

$F=mg-ma=m(g-a)$

$F=1.7\times 10^3(9.8-0.6)=1.56\times 10^4 N$

Hence, the tension in the cable during deceleration= $1.56\times 10^4 N$ .

d.We have to find that the height of elevator moved above its original staring point .

We have to find the final velocity.

$y_1=ut_1+\frac{1}{2}a_1t_1^2$

$y_1=0(1.50+\frac{1}{2}(1.2)(1.5)=1.35 m$

Final velocity

$v_1=a_1t_1$

$v_1=(1.2)(1.5)=1.8 m/s$

$y_2=v_1t_1=(1.8)(8.5)=15.3 m$

The third distance $y_3$ is up in the direction because the elevator decelerate upward which mean the distance is positive and is given by

$y_3=(1.8)(3)-\frac{1}{2}(0.6)(3)^2=2.7 m$

The total distance moved by elevator from its original position

$d=1.35+1.8+2.7=19.35 m$

Final velocity=0

5 0

3 years ago

When a sound from a source is refracted away from the surface of the earth, it gives an indication that the

White raven [17]

Answer:

the answer is A. air is cooler than the ground.

Explanation:

7 0

4 years ago

Read 2 more answers

PLEASE HELP!!! 50 points!!!!!! SEE ATTACHMENT!!! In which direction is there a net force of 200 N? left, right, up, down

Sladkaya [172]

The answer is right. The point and the 200 N sign maybe on the left side but the direction can be going right. I just took the quiz today, I wouldn't give the wrong answer cause i'm not a liar. -_-

5 0

3 years ago

Read 2 more answers

A coil of wire has been wrapped into a circle of 100 times with a diameter of 2.0 m. The coil is placed in a magnetic field of 5

jek_recluse [69]

Answer:

1570 V

Explanation:

N = 100

diameter = 2 m

radius = 1 m

B = 5 tesla

initailly the angle between the area vector of the coil and the magnetic field is zero and then the angle becomes 180 degree.

t = 2 s

Let Φ be the initial magnetic flux.

Φ = B A Cos 0 = 5 x 3.14 x 1 x 1 x 1 = 15.7 Wb

Let Φ' be the final magnetic flux

Φ' = B A Cos 180 = - 5 x 3,14 x 1 x 1 x 1 = - 15.7 Wb

According to the law of electromagnetic induction, the induced emf developed is equal to the rate of change of magnetic flux linked with the coil.

$e = -N\frac{d\phi }{dt}=\frac{\phi '-\phi }{t}$

$e =-100\frac{-15.7-15.7}{2}=1570 V$

4 0

3 years ago