The ratio of translational energy and total energy of a rolling sphere is given by

The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s isJ. (Formula: )

tekilochka [14]

25 Joule
Formula=.5*mass*velocity^2

8 0

3 years ago

Read 2 more answers

WILL MARK BRAINLIEST!!!!

zloy xaker [14]

Answer:

Wave model

Explanation:

7 0

3 years ago

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Kim is ice-skating going 4.6 m/s. What is her velocity after 10 seconds ?

MArishka [77]

This is a uniform rectilinear motion (MRU) exercise.

To start solving this exercise, we obtain the following data:

v = 4.6 m/s
d = ¿?
t = 10 sec

To calculate distance, speed is multiplied by time.

We apply the following formula: d = v * t.

We substitute the data in the formula: the <u>speed is equal to 4.6 m/s,</u> the <u>time is equal to 10 s</u>, which is left as follows:

$\bf{d=4.6\dfrac{m}{\not{s}}*10\not{s} }$

$\bf{d=46 \ m}$

Therefore, the speed at 10 seconds is 46 meters.

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6 0

2 years ago

A rectangular coil of wire (a = 22.0 cm, b = 46.0 cm) containing a single turn is placed in a uniform 4.60 T magnetic field, as

frez [133]

Explanation:

Below is an attachment containing the solution.

4 0

3 years ago

A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t

mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

$K=U$

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}$

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}$

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

$a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}$

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

$T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s$

So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

7 0

3 years ago