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Naya [18.7K]
3 years ago
5

Hi there hope your having a great day!! my questions both are SCIENCE laws of motion related fyi

Physics
1 answer:
Ann [662]3 years ago
4 0
1. 6.0 m

force/mass× acceleration is the equation

since the accelaration is unknown you would divide 180 by the mass of 30kg which is 6

2.A


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Marina86 [1]
To make sure the answer is correct
6 0
2 years ago
a 1000 khz am radio station broadcasts with a power of 20 kw. how many photons does the transmitter antenna emit each second
Gemiola [76]

1000 khz am radio station broadcasts with a power of 20 kw number of photon emitted per second is 30.16 x 10^30 photon/s.

The frequency of the radio station is:
f
=
1000
k
H
z
=
 1
×
10^6Hz
The transmit power is: P = 20kW = 20 X 10^3 W

The transmit power is: h = 6.63 x 10 ^-34 m^2.kg/s

The number of photon emitted per second = N = P/hf = <u>30.16 x 10^30 </u>photon/s.

1000 khz am radio station broadcasts with a photon of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw.1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw.


Learn more about photon on:
brainly.com/question/20912241

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6 0
1 year ago
A car is traveling on a straight, level road under wintry conditions. Seeing a patch of ice ahead of her, the driver of the car
dangina [55]
Her magnitude of deceleration on the ice would be 15.126m/s
5 0
2 years ago
The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be
bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

F=\frac{Gm_{1} m_{2} }{d^{2} }      ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }

d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}

d=\sqrt{1.48\times10^{17}}

d = 3.85 x 10⁸ m

3 0
3 years ago
Along the frictionless path you have chosen, the main force(s) acting on the puck after receiving the kick is (are):
Vaselesa [24]
Gravity, and Normal. Check the comments for why Applied isn't one.
5 0
3 years ago
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