Ask question

Ask question

All categories

3 years ago

12

When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The play

er then lands on the floor with their feet and is quickly brought to a stop. If the the floor deforms by 0.025 m while the player lands on it, what was the average force under the player's feet during the landing

1 answer:

saw5 [17]3 years ago

6 0

Answer:

2361.6N

Explanation:

Mass of player = 82kg

Velocity = 1.2m/s

Kinetic energy of player:

= 1/2mv²

= 1/2*82*1.2²

= 41x1.44

= 59.04J

Final kinetic energy = 0

Change in kinetic energy

|∆k| = |0-59.04|

= 59.04

Workdone by the feet = fd

d = 0.025

Fd = 59.04

F = 59.04/0.025

= 2361.6N

This is his average force.

You might be interested in

What are tectonic plates, and which of Earht's layers are they composed of?

motikmotik

The tectonic plates are made up of Earth's crust and the upper part of the mantle layer underneath. Together the crust and upper mantle are called the lithosphere. hope this helps :)

3 0

3 years ago

In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude

melamori03 [73]

Answers:
(a) $B_o = 0.3466$ μT
(b) $\lambda = 0.4488$ μm
(c) f = $6.68 * 10^{14}Hz$

Explanation:
Given electric field(in y direction) equation:
$E_y = 104sin(1.40 * 10^7 x -\omega t)$

(a) The amplitude of electric field is $E_o = 104$ . Hence

The amplitude of magnetic field oscillations is $B_o = \frac{E_o}{c}$
Where c = speed of light

Therefore,
$B_o = \frac{104}{3*10^8} = 0.3466$ μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
$\frac{2 \pi }{\lambda} = 1.40 * 10^7$
$\lambda = \frac{2 \pi}{1.40} * 10^{-7}$
$\lambda = 0.4488$ μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = $6.68 * 10^{14}Hz$

6 0

3 years ago

While in a car at 4.47 meters per second a passenger drops a ball from a height of 0.70 meters above the top of a bucket how far

viktelen [127]

Answer:

1.7 m

Explanation:

$v_x$ = Velocity of ball in x direction = 4.47 m/s

$u_y$ = Velocity of ball in y direction = 0

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

t = Time taken

$s_y$ = Vertical displacement = 0.7 m

$s_y=u_yt+\dfrac{1}{2}gt^2\\\Rightarrow 0.7=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{0.7\times 2}{9.81}}\\\Rightarrow t=0.38\ \text{s}$

Horizontal displacement is given by

$s_x=v_xt\\\Rightarrow s_x=4.47\times 0.38\\\Rightarrow s_x=1.7\ \text{m}$

The passenger should throw the ball 1.7 m in front of the bucket.

5 0

3 years ago

If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the

kodGreya [7K]

Answer:

a)

$v = 4.048 *10^6$ m/s

b)

Angular frequency = $1.92 * 10^7$

Explanation:

As we know

$v = \frac{qBr}{m}$

q is the charge on the electron = $3.2 * 10^{-19}$ C

B is the magnetic field in Tesla = $0.4$ T

r is the radius of the circle = $0.21$ m

mass of the electrons = $6.64 * 10^{-27}$ Kg

a)

Substituting the given values in above equation, we get -

$v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6$ m/s

b)

Angular frequency =

$\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7$

8 0

2 years ago

A plane flying horizontally at 377 m/s releases a package at an altitude of 15,770 m. How long will the package take to reach th

wolverine [178]

Answer:

An object is called a horizontal projectile if it is launched from a certain height with some initial horizontal velocity only. The initial vertical velocity of such an object is zero. But as the object falls through the atmosphere the horizontal component of velocity remains constant but vertical component increases due to gravitational acceleration.

Explanation:

3 0

3 years ago