Thermal energy transfers from a cup of tea at 350 K to the hand holding it. <br> A)True<br> B)False

A snowball is rolling down a hill at 4.5 m/s and accumulating snow as it goes. Its diameter begins at 0.50 m and ends at the bot

Reil [10]

To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.

The formula for centripetal acceleration is:
Centripetal Acceleration = v squared divided by r

where:
v = velocity, m/s
r= radium, m

assuming the velocity does not change:

at the top of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 0.25 m
= 81 m/s^2

at the bottom of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 1.25 m
= 16.2 m/s^2

to find the change in centripetal acceleration, take the difference of the two.
change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill

= 81 m/s^2 - 16.2 m/s^2
= 64.8 m/s^2 or 65 m/s^2

6 0

3 years ago

What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?

MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^2}$

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

$I = \frac{100}{4\pi (2.5)^2}$

$I = 1.2738W/m^2$

The relation between intensity I and $E_{max}$

$I = \frac{E_max^2}{2\mu_0 c}$

Here,

$\mu_0$ = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

$E_{max}^2 = 2I\mu_0 c$

$E_{max}=\sqrt{2I\mu_0 c }$

$E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)$

$E_{max} = 30.982 V/m$

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

$B_{max} = \frac{E_{max}}{c}$

$B_{max} = \frac{30.982 V /m}{3*10^8}$

$B_{max} = 1.03275 *10{-7} T$

Therefore the maximum value of the magnetic field is $B_{max} = 1.03275 *10{-7} T$

3 0

3 years ago

What should Ben wear to best protect himself during the experiment?

GarryVolchara [31]

Goggles, a lab coat, chemical gloves, and close- toed shoes

4 0

2 years ago

How can i prove the conservation of mechanical energy?

FinnZ [79.3K]

Answer:

We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved

Explanation:

5 0

2 years ago

A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5

harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

$E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\$

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

$P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\$

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0

2 years ago

Thermal energy transfers from a cup of tea at 350 K to the hand holding it. A)True B)False