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2 years ago

Thermal energy is the _____ of the kinetic and potential energies of all particles in an object.

Physics

1 answer:

Sphinxa [80]2 years ago

3 0

Thermal energy is the force of the kinetic and potential energies of all particles in an object

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Frequency is the ____________ that move past a point during a specific amount of time. Frequency is measured in ____________ , a

yulyashka [42]

Answer:

Frequency is the number of waves that move past a point during a specific amount of time. Frequency is measured in Hertz, and is classified as high, medium, or low. Frequency is interpreted as the pitch of a sound. Intensity refers to the loudness of a sound and is measured in decibels. Louder sounds increase the rate of nerve signals relayed to the brain.

Explanation:

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3 years ago

A car travels 6 m north for 2 seconds and 8 m south for 5 seconds. What is the car's

Degger [83]

2 m/s. 6+8/2+5 = 14/7 = 2

8 0

2 years ago

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What kind of eruption is likely to occur in a volcano having magma that is low in silica

Masja [62]

A volcano erupts quietly if it's magma is low on silica. Low silica magma will be very thin, runny, and will flow easily, so the gases in the magma will bubble out gently. Hope this helps!:)

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3 years ago

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Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22

natka813 [3]

Answer:

2420 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 22.9 N

Angle (θ) = 35°

Distance (d) = 129 m

Workdone (Wd) =?

The work done can be obtained by using the following formula:

Wd = Fd × Cos θ

Wd = 22.9 × 129 × Cos 35

Wd = 22.9 × 129 × 0.8192

Wd ≈ 2420 J

Thus, the workdone is 2420 J.

3 0

3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago