Ask question

Ask question

All categories

3 years ago

9

A rock is thrown downward from the top of a 36.2-m-tall tower with an initial speed of 13 m/s. Assuming negligible air resistanc

e, what is the speed of the rock just before hitting the ground

1 answer:

vekshin13 years ago

4 0

Explanation:

u = 13m/s

s = 36.2m

a = 9.81m/s² (positive because it is acting on the same direction as the velocity)

Using Kinematics, we have v² = u² + 2as = (13)² + 2(9.81)(36.2) = 879.244m²/s². => v = 29.65m/s.

You might be interested in

Mechanical energy is the sum of ________ energy and potential energy.apex

crimeas [40]

Mechanical energy is the sum of kinetic energy and potential energy

6 0

3 years ago

I need help ASAP. This is for 15 points

kramer

Answer:

Latitude :

runs: east to west

measures : distances north and south of the equator

Longitude :

runs : north to south

measures : the distance east or west of the Prime Meridian

7 0

3 years ago

if the 50 kg object slows down to a velocity of 1 m/s how much kinetic energy does it have 100j 50j 25j or none

rodikova [14]

$ek = \frac{1}{2}m {v}^{2}$

$ek = \frac{1}{2}(50) {(1)}^{2}$

$ek = \frac{1}{2}(50)$

$ek = 25j$

//

I'm not really sure but I do know that it's not 0 because the object is still moving, even if it's only moving at 1 m/s.

6 0

3 years ago

How many wavelengths are seen in this image of a sound wave?<br><br> three<br> four<br> six<br> ten

Anuta_ua [19.1K]

Four water levels are shown

7 0

3 years ago

Read 2 more answers

A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one

alekssr [168]

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R= $13\Omega$

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper wire=Sum of resistance of two piece of wires

$r+7r=13$

$8r=13$

$r=\frac{13}{8}$ ohm

Resistance of long piece of wire= $7\times\frac{13}{8}=\frac{91}{8}\Omega$

Resistance of short piece of wire = $\frac{13}{8}\Omega$

Resistivity of wire and cross section area of wire remains same .

Let L be the total length of wire and L' be the length of short piece of wire.

We know that

$R=\frac{\rho L}{A}$ = $\frac{\rho}{A}L=KL$

$\frac{R}{L}=K$

Where K= $\frac{\rho}{A}$ =Constant

Using the formula

$\frac{13}{L}=\frac{\frac{13}{8}}{L'}$

$\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}$

$L'=\frac{L}{8}$

Length of short piece of wire=L'= $\frac{L}{8}$

Length of long piece of wire= $L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L$

% of length of short piece of wire= $\frac{\frac{L}{8}}{L}\times 100=12.5%$ %

The resistance of the short piece= $\frac{13}{8}\Omega$

The resistance of the long piece= $\frac{91}{8}\Omega$

8 0

3 years ago