Temperature is a measure of

An object falls from rest on a high tower and takes 5.0 s to hit the ground. Calculate the object's position from the top of the

Lena [83]

Answer:

After 1 sec = 4.9 m

After 2 sec = 19.6 m

After 3 sec = 44.1 m

After 4 sec = 78.4 m

After 5 sec = 122.5 m

Explanation:

After 1 sec:

u=0m/s t=1 s a=9.8m/s²

s = ut + (1/2)at²

=0(1) + (1/2)(9.8)(1²) = 4.9m

After 2 sec:

u=0m/s t=2 s a=9.8m/s²

s = ut + (1/2)at²

=0(2) + (1/2)(9.8)(2²) = 19.6m

After 3 sec:

u=0m/s t=3 s a=9.8m/s²

s = ut + (1/2)at²

=0(3) + (1/2)(9.8)(3²) = 44.1m

After 4 sec:

u=0m/s t=4 s a=9.8m/s²

s = ut + (1/2)at²

=0(4) + (1/2)(9.8)(4²) = 78.4m

After 5 sec:

u=0m/s t=5 s a=9.8m/s²

s = ut + (1/2)at²

=0(5) + (1/2)(9.8)(5²) = 122.5m

7 0

3 years ago

Two coaxial conducting cylindrical shells have equal and opposite charges. The inner shell has charge +q and an outer radius a,

Leviafan [203]

Answer:

$\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}$

Explanation:

As we know that the charge per unit length of the long cylinder is given as

$\lambda = \frac{q}{L}$

here we know that the electric field between two cylinders is given by

$E = \frac{2k\lambda}{r}$

now we know that electric potential and electric field is related to each other as

$\Delta V = - \int E.dr$

$\Delta V = -\int_a^b (\frac{2k\lambda}{r})dr$

$\Delta V = -2k \lambda ln(\frac{b}{a})$

$\Delta V = \frac{\lambda ln(\frac{b}{a})}{2\pi \epsilon_0}$

$\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}$

7 0

3 years ago

How could you use a rock to measure the height of your ceiling?

Annette [7]

Answer:

I mean this is what I think

Explanation:

you would need to place a rock on top of each other until you reach the ceiling

It seems logical to me

7 0

3 years ago

What are organelles? Which part of a cell holds organelles?

DanielleElmas [232]

Answer:

An organelle is a structure in a cell that has specific jobs to do. Organelles are held in the cytoplasm

3 0

3 years ago

Read 2 more answers

A backyard swimming pool with a circular base of diameter 6.00m is filled to depth 1.50m. (b) Two persons with combined mass 150

Mashutka [201]

The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

$\rho \mathrm{Vg}=2 \mathrm{mg} \\$