All categories

3 years ago

Which formulas are used to calculate potential and kinetic energy

Physics

2 answers:

solniwko [45]3 years ago

8 0

Answer:

$\boxed{\bold { \large { \boxed {KE=\frac{1}{2} mv^2 \ , \ PE=mgh}}}}$

Explanation:

Kinetic energy formula

$\displaystyle KE=\frac{1}{2} mv^2$

Potential energy formula

$\displaystyle PE=mgh$

$\displaystyle KE \Rightarrow \sf kinetic \ energy \ (J)$

$\displaystyle PE \Rightarrow \sf potential \ energy \ (J)$

$\displaystyle m \Rightarrow \sf mass \ (kg)$

$\displaystyle v \Rightarrow \sf velocity \ (m/s)$

$\displaystyle g \Rightarrow \sf acceleration \ of \ gravity\ (m/s^2)$

$\displaystyle h \Rightarrow \sf height \ (m)$

aliya0001 [1]3 years ago

5 0

Answer:

PE=mgh

KE=<u>1</u> mv²

hope its helpful for u ..

You might be interested in

Two ice skaters, Lilly and John, face each other while at rest, and then push against each other's hands. The mass of John is tw

seropon [69]

Answer:

Lilly's speed is two times John's speed.

Explanation:

m = Mass

a = Acceleration

t = Time taken

u = Initial velocity

v = Final velocity

The force they apply on each other will be equal

$F=ma\\\Rightarrow a_l=\frac{F}{m_l}$

$F=ma\\\Rightarrow a_j=\frac{F}{2m_l}\\\Rightarrow a_j=\frac{1}{2}a_l$

$v=u+at\\\Rightarrow v_l=0+\frac{F}{m_l}\times t\\\Rightarrow v_l=a_lt$

$v=u+at\\\Rightarrow v_l=0+\frac{F}{2m_l}\times t\\\Rightarrow v_j=\frac{1}{2}a_lt\\\Rightarrow v_j=\frac{1}{2}v_l\\\Rightarrow v_l=2v_j$

Hence, Lilly's speed is two times John's speed.

4 0

3 years ago

If you weigh 400 N on the Earth, what would you weigh on the moon?

Alchen [17]

Answer:

14.87 pounds or 66.17 n

Explanation:

6 0

4 years ago

Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accele

lyudmila [28]

Answer:

Part a)

$v = 3.16 \times 10^7 m/s$

Part b)

r = 0.166 m

Explanation:

Part a)

As we know that the energy of the Hydride ion is given as

$E = 5 MeV$

here we have

$\frac{1}{2}mv^2 = 5\times 10^6(1.6 \times 10^{-19})$

also we know that

$m = 1.6 \times 10^{-27} kg$

now we have

$v = \sqrt{\frac{2 \times 5 \times 10^6(1.6 \times 10^{-19}}{1.6\times 10^{-27}}$

$v = 3.16 \times 10^7 m/s$

Part b)

As we know that magnetic force on the charge is centripetal force

so we have

$qvB = \frac{mv^2}{r}$

so we have

$r = \frac{mv}{qB}$

so we have

$r = \frac{1.6 \times 10^{-27}(3.16 \times 10^7)}{(1.6 \times 10^{-19}) 1.9}$

$r = 0.166 m$

4 0

3 years ago

A boat with an anchor on board floats in a swimming pool that is somewhat wider than the boat. Does the pool water level move up

pentagon [3]

Answer:

a) moves down

b) moves down

c) level remains same

Explanation:

Given that the anchor is initially on the floating boat.

In this condition initially the the volume of water $(V_w_i)$ displaced is to balance its weight.

Now,

$W_a=W_w$

$V_a.\rho_a.g=V_w_i.\rho_w.g$

$\frac{\rho_a}{\rho_w} =\frac{V_w_i}{V_a}$

We've, the density of steel $= 7850\ kg.^{-3}$ and the density of water $= 1000\ kg.^{-3}$

$\therefore V_w_i=7.85\times V_a$

When the anchor is dropped into water:

The volume of water displaced be $(V_w_f)$ which will be equal to the volume of anchor since it is immersed into it.

$V_a=V_w_f$

$\therefore V_w_i>V_w_f$ ...................(1)

So the level of water falls when the anchor is dropped into water.

Now, when the anchor is thrown on the ground the water has now less weight to balance so the water level falls down.

When the cork on the from the boat is dropped into the water and it still floats then it must displace same amount of water, hence there should be no change in the water level.

6 0

3 years ago

if a 60 kg person was standing on a platform at the surface of saturn and they jumped, they would have to push with a force grea

lidiya [134]

Answer:

A 60 kg person standing on a platform at the surface of Saturn and they jumped, they would have to push with a force greater than 540 N

Explanation:

The gravitational attraction between an object on the surface of a planet and the planet is given by the weight of the object

Therefore the force needed to be applied for an object to lift off the surface of a planet = The weight of the object

The weight of the object on the surface of a planet = m × g

Where;

m = The mass of the object

g = The strength of gravity on the planet's surface in N/kg

The given parameters are;

The mass of the person standing on a platform at the surface of Saturn, m = 60 kg

The strength of gravity on the surface of Saturn = 9 N/kg

Therefore, we have;

The weight of the person = The force greater than which the person would have to push on the surface of Saturn so as to Jump = The weight of the person on the surface of Saturn = 60 kg × 9 N/kg = 540 N

Therefore, for a 60 kg person standing on a platform at the surface of Saturn and they jumped, they would have to push with a force greater than 540 N.

4 0

3 years ago