Ask question

Ask question

All categories

3 years ago

11

A elephant kicks a 5.0\,\text {kg}5.0kg5, point, 0, start text, k, g, end text stone with 150\,\text J150J150, start text, J, en

d text of kinetic energy.
What is the stone’s speed?

1 answer:

S_A_V [24]3 years ago

3 0

The speed of the stone is 7.7 m/s

Explanation:

The kinetic energy of a body is the energy possessed by the body due to its motion. Mathematically,

$K=\frac{1}{2}mv^2$

where

m is the mass of the body

v is its speed

For the stone in this problem, we have:

K = 150 J is its kinetic energy

m = 5.0 kg is its mass

Re-arranging the equation for v, we find the speed of the stone:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(150)}{5.0}}=7.7 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

You might be interested in

10. A flashlight runs on a battery. If the light is left on the battery runs out and the flashlight

schepotkina [342]

Well I believe the claim is if you use all its energy fr a long period of time it does so the claim is if the light is left on the battery runs out and flashlight stops working

3 0

3 years ago

A 5 inch tall balloon shoot doubles in height every 3 days. if the equation y=ab^x, where is x is the number of doubling periods

VashaNatasha [74]

In this item if we let x be the number of doubling period, that is number of days divided by 3, and y be the height of the bamboo shoot, in the equation
y = ab^x
The value of a is 5 because that is the original height of the shoot and b is 2 because we are talking about the doubling rate. The equation is therefore,
y = 5(2^x)

3 0

3 years ago

If a typical smelt weighs 225 g, what is the total mass of pcbs in a smelt in the great lakes? If a typical smelt weighs 225 g,

photoshop1234 [79]

Answer;

= 2.34 x 10^-4 g pcbs

Explanation;

-A typical smelt has 1.04 ppm of PCB is great lakes.

Therefore;

The amount of PCB in a smelt in great leak can be calculated by multiplying the mass of smelt by the pcbs present in a smelt

= 225 g * ( 1.04/100000)

= 2.34 x10^-4 g pcbs

7 0

3 years ago

The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A

love history [14]

Answer:

Part a)

$EMF = 14 \times 10^{-3} V$

Part b)

$EMF = 15.67 \times 10^{-3} V$

Explanation:

As we know that magnetic flux through the loop is given as

$\phi = B.A$

now we have

$\phi = B\pi r^2$

now rate of change in flux is given as

$\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}$

now we know that

$A = \pi r^2$

$0.285 = \pi r^2$

$r = 0.30 m$

Now plug in all data

$EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)$

$EMF = 14 \times 10^{-3} V$

Part b)

Now the radius of the loop after t = 1 s

$r_1 = r_0 + \frac{dr}{dt}$

$r_1 = 0.30 + 0.037$

$r_1 = 0.337 m$

Now plug in data in above equation

$EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)$

$EMF = 15.67 \times 10^{-3} V$

5 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