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3 years ago

A 11,000-watt radio station transmits at 880 kHz. Determine the number of joules transmitted per second.

Physics

1 answer:

NARA [144]3 years ago

3 0

1 watt = 1 joule per sec

11,000 Watts = 11,000 joules per sec

The frequency doesn't matter.

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Find the magnitude and direction of an electric field that exerts a 4.80×10−17N westward force on an electron. (b) What magnitud

jok3333 [9.3K]

Explanation:

(a) E = F/q

E = 4.8×10^-17/1.6×10^-19

E = 300 N/C

(b) same magnitude of electric field is exerted on proton

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3 years ago

An astronaut on Pluto attaches a small brass ball to a 1.00-m length of string and makes a simple pendulum. She times 10 complet

pickupchik [31]

Answer:

The acceleration due to gravity at Pluto is 0.0597 m/s^2.

Explanation:

Length, L = 1 m

10 oscillations in 257 seconds

Time period, T = 257/10 = 25.7 s

Let the acceleration due to gravity is g.

Use the formula of time period of simple pendulum

$T = 2\pi\sqrt{\frac{L}{g}}\\\\25.7 = 2 \times 31.4\sqrt{\frac{1}{g}}\\\\g = 0.0597 m/s^2$

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3 years ago

A child pushes a toy box across the floor in 5 seconds. If he did 30 J of work on the toy box, what amount of power was required

s2008m [1.1K]

My answer is 6 watts because 30J/5s is 6

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3 years ago

Read 2 more answers

The gravitational force between two objects will be greatest in which of the following situations?

Kobotan [32]

Answer:

Explanation:

Gravitational law states that, the force of attraction or repulsion between two masses is directly proportional to the product of the two masses and inversely proportional to the square of their distance apart.

So,

Let the masses be M1 and M2,

F ∝ M1 × M2

Let the distance apart be R

F ∝ 1 / R²

Combining the two equation

F ∝ M1•M2 / R²

G is the constant of proportional and it is called gravitational constant

F = G•M1•M2 / R²

So, to increase the gravitational force, the masses to the object must be increased and the distance apart must be reduced.

So, option c is correct

C. Both objects have large masses and are close together.

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3 years ago

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for

uysha [10]

Answer:

F=5449 N

Explanation:

Work done is a product of force and displacement ie

Work done, W, = Force*Displacement

Power, P, is Work done/Time

$P=\frac {W}{t}=\frac {FS}{t}$ where P is power, W is work done, F is force, S is displacement and t is time

In this case, F is the frictional force. Converting the power from hp to W, we multiply by 746 hence P=746*168=125328 W

Since displacement/time is velocity, then

P=FV where V is velocity in m/s

Making F the subject

$F=\frac {P}{V}$

$F=\frac {125328}{23}=5449.043478 N$

F=5449 N

7 0

3 years ago