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Likurg_2 [28]
3 years ago
12

Which describes why people on earth can see light from the stars in the sky that are so far away?

Physics
1 answer:
balandron [24]3 years ago
3 0
Choice-C is a correct statement. 

Each of the other choices combines elements of truth
with elements of hogwash.
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If the force used to push a shopping cart increases, the cart's acceleration will
rewona [7]

Answer:

b) increase

Explanation:

4 0
3 years ago
Read 2 more answers
A Carnot engine has an efficiency of 83.0% and performs 4500 J of work every cycle. How much energy is discharged to the lower t
ololo11 [35]

We will start by finding the heat in state one through the thermal efficiency or efficiency of a thermal machine is a coefficient or dimensionless ratio calculated as the ratio of the energy produced (in an operating cycle) and the energy supplied to the machine. Mathematically it is the relationship between the work generated and the heat emanated.

So,

\eta = \frac{W}{Q_1} \rightarrow Q_1 = \frac{W}{\eta}

Q_1 = \frac{4500}{0.83}

Q_1 = 5421J

The total change of energy is equivalent to 4500J and this is equal by conservation of energy to the total change in heat. So:

W = \Delta Q

W = Q_1-Q_2

4500 = 5421- Q_2

Q_2 = 921.68 J

Therefore the energy which is discharged to the lower temperature reservoir every cicle is 921.7J

6 0
2 years ago
Part II # 1 A mass on a string of unknown length oscillates as a pendulum with a period of 4 sec. What is the period if: (Parts
Mrac [35]

Answer:

C. 1) No Change (4 sec) 2) 5.7 sec 3) 2.8 sec 4) No Change (4 sec)

Explanation:

Given that:

Period (T) = 4 s

1) If the mass is doubled.

The period of a pendulum is given by the formula:

T=2\pi\sqrt{\frac{L}{g} } where L is the length and g is the acceleration due to gravity.

From the formula, the period does not depend on the mass of the spring therefore if the mass is doubled the period does not change.

2) The string length is doubled

Given that:

T=2\pi\sqrt{\frac{L}{g} }, but\ T =4s\\4=2\pi\sqrt{\frac{L}{g} }

if the length is doubled, the new spring length is 2L. Therefore the new period (T1) is given as:

T_1=2\pi\sqrt{\frac{2L}{g} }=\sqrt{2} (2\pi\sqrt{\frac{L}{g} })=\sqrt{2}*4=5.7\ sec

3) The string length is halved

Given that:

T=2\pi\sqrt{\frac{L}{g} }, but\ T =4s\\4=2\pi\sqrt{\frac{L}{g} }

if the length is halved, the new spring length is L/2. Therefore the new period (T1) is given as:

T_1=2\pi\sqrt{\frac{L}{2g} }=\sqrt{1/2} (2\pi\sqrt{\frac{L}{g} })=\sqrt{1/2}*4=2.8\ sec

4) The amplitude is halved

From the formula, the period does not depend on the amplitude therefore if the amplitude is halved the period does not change.

6 0
3 years ago
A long solenoid that has 1,140 turns uniformly distributed over a length of 0.350 m produces a magnetic field of magnitude 1.00
Troyanec [42]

Answer:

I = 2.4 *10^{-2} A = 2.4 mA

Explanation:

The magnetic field B inside long solenoid with current I is given as

B = \frac{N \mu_o I}{L}

where

N is number of turn of solenoids = 1140 turns

\mu_0 = 4*\pi *10^{-7} T.m

I is current that passes through solenoid

L is length along which current pass = 0.350 m

plugging value to get required value of current

1.00*10^{-4} = \frac{1140*4*\pi *10^{-7} I}{0.350}

I = 2.4 *10^{-2} A = 2.4 mA

3 0
3 years ago
Two loudspeakers (A and B) are 3.20m apart and emitting a sound with a frequency of 400Hz. An observer is 2.10m directly in fron
TiliK225 [7]

Answer:

The observer hears a loud sound

Explanation:

In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.

You first calculate the distance between the observer and the loudspeakers.

The distances are given by:

d1: distance to loudspeaker A = 2.10m

d2: distance to loudspeaker B

d_2=\sqrt{(3.20m)^2+(2.10m)^2}=3.827m

Next, you calculate the wavelength of the sound waves by using the following formula:

\lambda=\frac{v_s}{f}

vs: speed of sound =  343 m/s

f: frequency of the waves = 400Hz

λ: wavelength

\lambda=\frac{343m/s}{400Hz}=0.8575m

Next, you calculate the path difference between the distance from the observer to the loudspeakers:

\Delta d=3.827m-2.10m=1.727m

You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:

\frac{\Delta d}{\lambda}=\frac{1.727m}{0.857}\approx2

Then, there will be a constructive interference, and the sound who the observer hears is loud.

5 0
2 years ago
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