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gtnhenbr [62]
3 years ago
6

Elias serves a volleyball at a velocity of 16 m/s. The mass of the volleyball is 0.27 kg. What is the height of the volleyball a

bove the gym floor when its total mechanical energy is 41.70 J? Round to the nearest tenth\
Physics
2 answers:
evablogger [386]3 years ago
7 0

Ingenuity says that the answer is 2.7 m

givi [52]3 years ago
4 0
Hello! The anwser is 2.7! I hope this helps! ^^
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Which of the following would be considered a "point source" of water
KATRIN_1 [288]

Answer:

c. Groundwater contamination at a fracking site

Explanation:

all others could be sources from tens of square kilometers of surface area.

Fracking is limited to within a short range of the well hole.

5 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
Which statement bet explains the relationship between the electric force between two charged objects and the distance between th
Nataliya [291]
Coulomb's Law: Force = k x q1x q2 divided distance square
where k=9x10^9 , q1 and q2 are the charge
So if you distance is halved, your force is stronger by 4 times
and if you distance is doubled, your force is 1/4
Ask me again if you aren't clear :)

4 0
3 years ago
I was walking with Ruby in a garden from rest in a straight line with uniform acceleration so that we covered 0.25 m in the fift
Serhud [2]

Answer:

0.02m/s^2

Explanation:

7 0
2 years ago
URGENT!!! sorry this is my first time using brainly! but here is my question: Two ropes are attached to a wagon, one horizontal
Elan Coil [88]

The components of the net force on the cart is determined as 67.66 N.

<h3>Component of net force on the cart</h3>

The component of net force on the cart is determined by resolving the forces into x and y -components.

T1 = 30 N

T2 = 40 N

T1x = -30cos(0) = 30 N

T1y = 30sin(0) = 0

T2x = 40 x cos(30) = 34.64 N

T2y = 40 x sin(3) = 20 N

∑X = 30 N + 34.64 N = 64.64 N

∑Y = 0 + 20 N = 20 N

<h3>Resultant force</h3>

R = √(64.64² + 20²)

R = 67.66 N

Learn more about net force here: brainly.com/question/25239010

#SPJ1

3 0
2 years ago
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