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3 years ago

3. Applying Concepts Explain why a canoe that has hollow, water-

Physics

1 answer:

Alekssandra [29.7K]3 years ago

5 0

Answer:

I may be wrong

Explanation:

it it won't collapse because it is like a log logs don't sink when they are in water

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Which of the following statements are true of thermal energy and kinetic

Elis [28]

Answer:

all the molecules or atoms in motion has kinetic energy

4 0

1 year ago

A charge of 80 coulombs passes through a circuit in 5 seconds. What is the current through the circuit?

UkoKoshka [18]

Answer:

I = 16amp

Explanation:

Charge coulomb ( Q ) = It

Where I =current in ampere

t = time = 5 seconds

80 = I × 5

I = 80/5

I = 16amp

The current through the circuit will be I = 16amp

4 0

3 years ago

Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an

bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

$\frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}}$ ......... (1)

Hence, total charge will be as follows.

$q_{1} + q_{2}$ = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

$q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}$

and, $q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}$

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

$q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}$

= $\frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}$

= 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

5 0

3 years ago

A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

$\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s$

$\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM$

Therefore the speed in RPM will be 62.64 RPM.

5 0

3 years ago

Which of the following is an example of an electromagnetic wave ?

Alex

Answer:

Its a wave

Explanation:

THis is one

3 0

3 years ago

Read 2 more answers