All categories

4 years ago

PLEASE HURRY WILL MARK BRAINLIEST IF CORRECT

Physics

1 answer:

Readme [11.4K]4 years ago

6 0

The answer to the question is 1.48 x 10^-11 J

You might be interested in

3. A rescue plane drops a package to a stranded party of explorers. The plane is traveling horizontally at a velocity of

Otrada [13]

Hi there!

For this part, we can disregard the horizontal velocity because horizontal and vertical motion are INDEPENDENT.

We can used the simplified equation to solve for time:

$t = \sqrt{\frac{2h}{g}}$

Where:

h = height

g = acceleration due to gravity (assume g = 10 m/s²)

Plug in given values:

$t = \sqrt{\frac{2(50)}{10}} = \sqrt{10}$

t ≈3.162 sec

3 0

3 years ago

A 5 kilogram cat is resting on top of a bookshelf that is 3 meters high. What is the cat’s gravitational potential energy relati

Free_Kalibri [48]

U=mgh
U=5kg*9.8N/kg*3m
U=147J

5 0

3 years ago

What is the pathway of sound through fluids starting at the oval window through to dissipation of the sound waves at the round w

ruslelena [56]

Perilymph of scala vestibule; endolymph of cochlear duct; perilymph of scala tympani

6 0

3 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

3 years ago

Water is considered to be

Natalka [10]

Answer: the thing that brings you life- h2o- idrk the question but if this helps your welcome ^-^have a good day

Explanation:

3 0

3 years ago