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Fiesta28 [93]
3 years ago
10

Calculate the impulse of a 1kg box that starts from rest and accelerates to 50m/s over a period of 10 seconds.

Physics
1 answer:
fomenos3 years ago
8 0

Answer:

please find attached pdf

Explanation:

Download pdf
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The correct answer to your question is 3.62
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What Could be the control in an experiment to measure the effects of gas additives on fuel
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Well,

A control in an experiment would basically be the "normal" version of your test subjects.

In a drug testing experiment with people, the control group would be the people who don't take the drug.

In an experiment on the effects of salt on potatoes, the control group would be a potato without salt on it.

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3 years ago
Suppose you are asked to find the amount of time t, in seconds, it takes for the turntable to reach its final rotational speed.
Annette [7]

Answer:

option (D)

Explanation:

Here initial rotation speed is given, final rotation speed is given and asking for time.

If we use

A) θ=θ0+ω0t+(1/2)αt2

For this equation, we don't have any information about the value of angular displacement and angular acceleration, so it is not useful.

B) ω=ω0+αt

For this equation, we don't have any information about angular acceleration, so it is not useful.

C) ω2=ω02+2α(θ−θ0)

In this equation, time is not included, so it is not useful.

D) So, more information is needed.

Thus, option (D) is true.

5 0
3 years ago
A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of
SCORPION-xisa [38]

Answer:

a).β=0.53x10^{-3} T

a).β=0.40 x10^{-4} T

Explanation:

The magnetic field at distance 'r' from the center of toroid is given by:

\beta =\frac{u_{o}*I*N}{2\pi*r}

a).

N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A}  \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T

b).

The distance is the radius add the cross section so:

r_{1}=15cm+5cm\\r_{1}=20cm

r_{1} =20cm*\frac{1m}{100cm}=0.20m

\beta =\frac{u_{o}*I*N}{2\pi*r1}

\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T

3 0
3 years ago
3. What are the challenges of looking for Dyson spheres?
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1. it is difficult to search for it . Because infrared rays will never penetrate through earth atmosphere.

2. we are unaware of how it looks like and we only know it is red and will glow . A damaged star also looks like this.

3. Dust also makes is hard to detect Dyson spheres . So we will get confused between Dyson sphere and a star surrounded by dust.

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