If she has a mass of 45kg and is 25 meters high, what is her potential energy?

HELP PLS! :/

zlopas [31]

Explanation:

formula: Vi = Vf - (at)

Vi: intial velocity

Vf: final velocity

a: acceleration

t: time

fill in formula with the numbers you are given

Vi= 41.6m - ((9.81 m/s^2)(1.89s))

parenthesis first according to pemdas

Vi= 41.6m - 18.54m/s

Answer: 23.06m/s

disclaimer: I havent done physics in awhile so I have no idea if this is right. just an attempt to help steer you in the right direction hopefully. good luck

7 0

3 years ago

Q = cmAT

castortr0y [4]

Answer: $Q=3000 cal$

Explanation:

We are given the following formula:

$Q=m. c. \Delta T$ (1)

Where:

$Q=3000 cal$ is the amount of heat

$m=300g$ is the mass of water

$c=1 cal/g \°C$ is the specific heat of water

$\Delta T$ is the variation in temperature, which in this case is $\Delta T=30\°C-20\°C=10\°C$

Rewriting equation (1) with the known values at the right side, we will prove the result is $3000 cal$ :

$Q=(300g)(1 cal/g \°C)(10\°C)$ (2)

$Q=3000 cal$ This is the result

8 0

3 years ago

Mass m is fired to the right at a speed of 7.0 m/s where it collides and becomes imbedded in mass M which is hanging from a rope

Tanya [424]

Answer:

V=0.28m/s, the combined mass will be 0.28m/s fast

h=0.00399m

Explanation:

Mass m is fired to the right at a speed of 7.0 m/s where it collides and becomes imbedded in mass M which is hanging from a rope. Mass m is 0.50kg and mass M is 12.0kg. a- How fast will the combined mass (m+M) be moving just after the collision? b- What is the maximum height the system (m+M) will reach?

step by step explanation:

from the principles of linear momentum which states that the total momentum before collision is equal to the total momentum after collision

momentum before collision=momentum after collision

notice that in this this case the collision is inelastic, owing to the fact that the two bodies stuck together after collision

mu+Mu2=(m+M)V

V=common velocity after collision

m=mass of the body fired

M=Mass hanging on the rope

u=velocity of the body

u2=velocity of the mass hanging on arope 0m/s

0.5*7+0=(0.5+12)V

3.5=12.5V

V=3.5/12.5

V=0.28m/s, the combined mass will be 0.28m/s fast

b. the maximum height reached after the impact

starting with the swing of the pendulum just after the collision until it reaches its maximum height. Conservation of mechanical energy applies here, so:

kinetic energy =potential energy

½ (m+M)V = (m+M)gh.

So, the speed of the system immediately after the collision is:

V = (2gh)½

0.28=(2*9.81*h)^0.5

0.0784=2*9.81h

h=0.00399m

8 0

4 years ago

____ are formed where bumps from two surfaces come into contact.

NemiM [27]

the answer is Friction.

hope this heslp

6 0

4 years ago

Read 2 more answers

A 550 N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plu

mario62 [17]

Answer:

(a) The elevator has an acceleration of 2.14 m/s² and direction is downward

(b) The acceleration is a=2.14 m/s²

Explanation:

Let +y be upward direction

For Part (a)

The mass of student (whose weight is 550N) is:

ms=w/g

ms=550/9.8=56.1 kg

Apply ∑Fy=ma

$n-m_{s}g=m_{s}a\\$

Solve for acceleration a

$a=\frac{n-m_{s}g}{m_{s}}\\ a=\frac{430N-550N}{56.1kg} \\a=-2.14m/s^{2}$

The elevator has an acceleration of 2.14 m/s² and direction is downward

Part(b)

For the acceleration is the scale reads 670 N

For n=670N

So

$a=\frac{n-m_{s}g}{m_{s}}\\ a=\frac{670N-550N}{56.1kg}\\ a=2.14m/s^{2}$

3 0

4 years ago