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Serggg [28]
3 years ago
6

A 10 kg mass rests on a table. What acceleration will be generated when a force of 5 N is applied? a 0.5 m/s2 b 3.5 m/s2 c 5 m/s

2 d 35 m/s2
Physics
1 answer:
rosijanka [135]3 years ago
3 0

Answer:

a= 0.5m/s^2

Explanation:

Force applied on an object is known as

F=m.a  (Newton's second law states it)

a=F/m

a=5/10=0.5m/s^2

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An L-R-C series circuit, R = 160 Ω , L = 0.790 H , and C = 1.30×10−2 μF . The source has a voltage amplitude of 140 V and a freq
wolverine [178]

Answer: a) 1 b) 61 W c) 61 W

Explanation:

a) The  Power Factor (also known as cos φ), is defined by the difference in phase between current and voltage, in a RLC series circuit, and is expressed as follows:

cos φ = R / Z = R / \sqrt{(R)^{2} + (Xl -Xc)^{2} }

In resonance, XL =XC, so the circuit behaves like it were purely resistive, so Z=R.

Replacing in the expression for power factor, we have:

cos φ = R/Z = R/R = 1

This means that in resonance, current and voltage are in phase each other.

b) The average power delivered by the source, in resonance, is simply the power dissipated at the resistance R, as follows:

Pavg = V² rms / R = V₀² / √2 / R = 61 W

c) If the circuit remains in resonance, the average power , which does not depends on frequency provided this condition remains, keeps the same, i.e. , 61 W.

7 0
3 years ago
A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0
nikklg [1K]

Answer:

(a) 10 s

(b) 30 m/s

(c) 150 m

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x = 15t

The officer's position at time t is:

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15t = 1.5 t²

t = 0 or 10 s

(b) The officer's speed is:

v = 3t

v = 30 m/s

(c) The position is:

x = 15t = 150 m

6 0
3 years ago
How many Mars-sized planets can fit in Jupiter
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Answer:

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Explanation:

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What is the half-life of a radioisotope
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4 0
3 years ago
Read 2 more answers
Estimate the length of the neptunian year using the fact that the earth is 1.50Ã108km from the sun on average.
Ksenya-84 [330]

Answer:

Explanation:

We know that,

Neptune is 4.5×10^9 km from the sun

And given that,

Earth is 1.5×10^8km from sun

Then,

Let P be the orbital period and

Let a be the semi-major axis

Using Keplers third law

Then, the relation between the orbital period and the semi major axis is

P² ∝ a³

Then,

P² = ka³

P²/a³ = k

So,

P(earth)²/a(earth)³ = P(neptune)² / a(neptune)³

Period of earth P(earth) =1year

Semi major axis of earth is

a(earth) = 1.5×10^8km

The semi major axis of Neptune is

a (Neptune) = 4.5×10^9km

So,

P(E)²/a(E)³ = P(N)² / a(N)³

1² / (1.5×10^8)³ = P(N)² / (4.5×10^9)³

Cross multiply

P(N)² = (4.5×10^9)³ / (1.5×10^8)³

P(N)² = 27000

P(N) =√27000

P(N) = 164.32years

The period of Neptune is 164.32years

6 0
3 years ago
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