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3 years ago

A foot player runs 1.6m/s and has a KE of 790 J. What is his mass?

1 answer:

3 0

The equation for kinetic energy is,

Ke = (1/2)mv^2.

You're given a kinetic energy of 790 joules, and a speed of 1.6 m/s. Plugging these values into the equation, we get,

790 = (1/2)(1.6)^2(m).

Solving for m, we get,

m = (790)/(0.5(1.6)^2).

I'll let you crunch out those numbers for yourself :D

If you have any questions, feel free to ask. Hope this helps!

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A super ball is dropped from a height of 100 feet. Each time it bounces, it rebounds half the distance it falls. How many feet w

Crank

The total distance travelled by the ball after the fourth impact is 275 feet.

<u>Explanation:</u>

Given-

Height, h = 100 feet

Rebounds half the distance

Distance in feet for the fourth time, x = ?

For the first time, the distance travelled by the ball is, x = 100 feet

For the second time, it will bounce up to 50 feet and fall upto 50 feet( half of 100 feet)

So, the distance travelled after the second impact, x = 100 + 50 + 50 = 200 feet

For the third time, it will bounce up to 25 feet and fall upto 25 feet( half of 50 feet)

So, the distance travelled after the third impact, x = 200 + 25 + 25 = 250 feet

For the fourth time, it will bounce up to 12.5 feet and fall upto 12.5 feet( half of 25 feet)

So, the distance travelled after the fourth impact, x = 250 + 12.5 + 12.5 = 275 feet

Therefore, total distance travelled by the ball after the fourth impact is 275 feet.

4 0

4 years ago

A bicycle is traveling north at 5.0 m/s. The mass of the wheel, 2.0 kg, is uniformly distributed along the rim, which has a radi

andreev551 [17]

Answer:2

Explanation:

Given

Velocity of bicycle is 5 m/s towards north

radius of rim $r=20 cm$

mass of rim $m=2 kg$

Angular momentum $\vec{L}=I\cdot \vec{\omega }$

$I=mr^2=2\times 0.2^2=0.08 kg-m^2$

$\omega =\frac{v}{r}=\frac{5}{0.2}=25 rad/s$

$L=0.08\times 25=2kg-m^2/s$

direction of Angular momentum will be towards west

7 0

3 years ago

The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally

dalvyx [7]

(a) The electrons move horizontally from west to east, while the magnetic field is directed downward, toward the surface. We can determine the direction of the force on the electron by using the right-hand rule:
- index finger: velocity --> due east
- middle finger: magnetic field --> downward
- thumb: force --> due north
However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
$K= \frac{1}{2}mv^2$
where K is the kinetic energy, m the electron mass and v their velocity. Re-arranging the formula, we find
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 2.20 \cdot 10^{-15} J}{9.1 \cdot 10^{-31} kg} }=6.95 \cdot 10^7 m/s$

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
$qvB = m \frac{v^2}{r}$
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
$r= \frac{mv}{qB}= \frac{(9.1 \cdot 10^{-31} kg)(6.95 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19} C)(3.00 \cdot 10^{-5} T)}=13.18 m$

And finally we can calculate the centripetal acceleration, given by:
$a_c = \frac{v^2}{r}= \frac{(6.95 \cdot 10^7 m/s)^2}{13.18 m}=3.66 \cdot 10^{14} m/s^2$

5 0

3 years ago

An archer shoots an arrow into a target using a bow. If the "action" force is the force of the bowstring against the arrow, what

aliya0001 [1]

Weight of an arrow is 20 plus 3

8 0

3 years ago

The distance from location C to location E, ________ to __________, equals one wavelength.

RSB [31]

Compression to compression answer would be c hope this helps

7 0

3 years ago

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