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3 years ago

Mechanical waves can travel through _______.

Physics

1 answer:

eimsori [14]3 years ago

5 0

mediums only, mechanical waves always propagate through medium

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a truck is moving around a circular curve @ a uniform velocity of 13m/s. if the centifrical force on truck is 3,300N with truck

VMariaS [17]

Answer

81.94 m

Explanation

The centripetal force of an object moving in a circular path is given by:

F = mv²/r Where m is the mass of the object, v is the constant velocity and r is the radius of the curve.

F = mv²/r

3,300 = (1600×13²)/r

3,300 = 270,400/r

r = 270,400/3,300

= 81.94 m

8 0

3 years ago

A farm tractor tows a 3300-kg trailer up a 14" incline with a steady speed of 2.8 m/s. what force does the tractor exert on the

gladu [14]

The force exerted by gravity is:

F = m g

F = 3300 kg * 9.8 m/s^2

F = 32,430 N

The force exerted due to the inclined plane is:

F tractor = 32,430 N * sin 14

F tractor = 7,823.75 N (answer)

7 0

3 years ago

What is the mass of an object moving at a speed of 10.0 m/s with a kinetic energy of 760 J?

rewona [7]

Answer:

Explanation:

3 0

3 years ago

A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 Î¼C on each plate. The plates have a

butalik [34]

Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance $C=260\ pF$

Charge $q=0.155\ \mu\ C$

Separation of plates = 0.313 mm

We need to calculate the potential difference between the plates

Using formula of potential difference

$V= \dfrac{Q}{C}$

Where, Q = charge

C = capacitance

Put the value into the formula

$V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}$

$V=596.2\ volts$

Hence,The potential difference between the plates is 596.2 volts.

7 0

4 years ago

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va

bonufazy [111]

Answer:

10000 V

0.00225988700565 m²

$8\times 10^{-12}\ F$

Explanation:

E = Electric field = $4\times 10^6\ V/m$

d = Gap = 2.5 mm

Q = Charge = 80 nC

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

Potential difference is given by

$V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V$

The potential difference between the plates is 10000 V

Area is given by

$A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2$

The area of the plate is 0.00225988700565 m²

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F$

The capacitance is $8\times 10^{-12}\ F$

4 0

3 years ago