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eduard
3 years ago
10

5 uses of mechanical energy​

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
8 0

Answer: The 5 uses of mechanical energy are: 1) Hammering a nail

2)  Using Dart gun 3) Moon 4) Hydropower plant 5) Sharping a pencil.

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As speed increases, how does the potential, kinetic, and total energy levels change?
Anastaziya [24]
Potential equals kenecric at the bottom so potential would also increas
4 0
3 years ago
Total these measurements. Your answer should indicate the proper accuracy. Be sure to include the units in your answer. (Remembe
vivado [14]

Answer:

10

Explanation:

This is tough. The last number  0.2 has only one significant figure. So while the sum of all the numbers is 12.3, you must only leave one sig figure. Rounding to the tenths gives 10.  

3 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
A wave travels at 175 m/s along the x-axis.If the period of the periodic vibrations of the wave is 3.00 milliseconds,then what i
OlgaM077 [116]

Answer:  

<h2>E) 52.5 cm</h2>

Explanation:

Step one:

given data

period T= 3 milliseconds= 0.003

velocity v= 175m/s

wave lenght λ=?

Step two:

we know that f=1/T

the expression relating period and wave lenght is

v=λ/T

λ=v*T

λ=175*0.002

λ=0.525m

to cm= 0.525*100

=52.5cm

The wavelength of the wave is E) 52.5 cm

7 0
2 years ago
Need help ASAP
kati45 [8]

Answer:

Explanation:

1.electric

2.radio wave

3.broadcast

Hope it helped you.

8 0
2 years ago
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