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levacccp [35]
3 years ago
11

a car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achiev

e this speed?
Physics
1 answer:
atroni [7]3 years ago
6 0
T=Vf-Vi/s
25m/s -15m/s/ 125m
10m/s /125m
=0.08s
I hope it’s correct !
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Match the theory to the statement that best describes it. 1. big bang. 2. steady state. 3.osscillating universe. 4. inflation Ch
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Explanation :

(1) Big bang : (1) The most accepted theory on the origin of the universe.

This theory shows the expanding of the universe from high density and high-temperature states.

(2) Steady state : (3) All is the same and will always stay the same.

Steady state means that the properties of any system remain the same always.

(3) Oscillating universe : (4) Agrees with the big bang theory but insists the universe expanded much quicker.

Oscillating universe theory is the result of big bang theory.

(4) Inflation Choices : (2) it's like an inflating and deflating balloon that never stops.  

In cosmology, cosmic inflation or deflation is just the expanding and contraction of the universe.

So, the statements and the choices are related as:

               (1)-(1)

              (2)- (3)

               (3)-(4)

               (4)-(2)

8 0
4 years ago
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A potential difference of 300 volts is applied to a 2.0-µf capacitor and an 8.0-µf capacitor connected in series. (a) What are t
zubka84 [21]

Answer:

a) Q1=Q2=480μC   V1=240V   V2=60V

b) Q1=96μC   Q2=384μC   V1=V2=48V

c) Q1=Q2=0C    V1=V2=0V

Explanation:

Let C1 = 2μC  and  C2=8μC

For part (a) of this problem, we know that charge in a series circuit, is the same in C1 and C2. Having this in mind, we can calculate equivalent capacitance first:

C=\frac{1}{\frac{1}{C1} +\frac{1}{C2} } = 1.6\mu C

Q_{T} = V_{T}*C_{T} = 480\mu C

V1 = \frac{Q1}{C1}=240V

V2 = \frac{Q2}{C2}=60V

For part (b), the capacitors are in parallel now. In this condition, the voltage is the same for both capacitors:

V1=V2   So,  \frac{Q1}{C1} = \frac{Q2}{C2}

Total charge is the same calculated for part (a), so:

\frac{Qt - Q2}{C1} = \frac{Q2}{C2}   Solving for Q2:

Q2 = 384μC    Q1 = 96μC.

Therefore:

V1=V2=48V

For part (c), both capacitors would discharge, since their total voltage of 300V would by applied to a wire (R=0Ω). There would flow a huge amount of current for a short period of time, and capacitors would be completely discharged. Q1=Q2=0C  V1=V2=0V

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Ultraviolet rays is the answer

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