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PIT_PIT [208]
3 years ago
13

Quincia is traveling at a speed of 35 m/s sees a squirrel run across the road in front of her and slams on the brakes. Four seco

nds later she comes to a complete stop. What is her acceleration?
Physics
1 answer:
kotegsom [21]3 years ago
8 0
Her <span>acceleration is 8.75 hope is helps</span>
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A substance with a define shape and volume is a
Vlad1618 [11]
Solid is the answer.
8 0
3 years ago
A world-class sprinter can burst out of the blocks to essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race.
wolverine [178]
Given:
u = 0, initial speed (sprinter starts from rest)
v = 11.5 m/s, final speed
s = 15 m, distance traveled to attain final speed.

Let
a =  average acceleration,
t = time taken to attain final speed.

Then
v² = u² + 2as
or
(11.5 m/s)² = 2*(a m/s²)*(15 m)
a = 11.5²/(2*15) = 4.408 m/s²

Also
v = u +a t
or
(11.5 m/s) = (4.408 m/s²)*(t s)
t = 11.5/4.408 = 2.609 s

Answer:
The average acceleration is 4.41 m/s² (nearest hundredth).
The time required is 2.61 s (nearest hundredth).
8 0
3 years ago
!!!!!!!!!!!PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!
Alexeev081 [22]
You might want to do a re-erp on this equation
5 0
3 years ago
Read 2 more answers
A Carnot engine operates between 1350 °F and 125 °F. If it rejects 55 Btu as heat, determine the work output.
ruslelena [56]

Answer:

C. W = 115.12\,Btu

Explanation:

Thermodynamically speaking, a Carnot engine represents an entirely reversible thermal process and its energy efficiency represents the maximum theoretical efficiency that thermal machines can reach. The efficiency of the ideal thermal process (\eta), no unit, is:

\eta = \left(1-\frac{T_{L}}{T_{H}} \right) (1)

Where:

T_{L} - Temperature of the cold reservoir, measured in Rankine.

T_{H} - Temperature of the hot reservoir, measured in Rankine.

If we know that T_{H} = 1809.67\,R and T_{L} = 584.67\,R, then the energy efficiency of the ideal thermal process is:

\eta = 0.678

By First Law of Thermodynamics, we calculate the work output:

W = Q_{H}-Q_{L}

W = \frac{W}{\eta} -Q_{L} (By definition of efficiency)

Q_{L} = \frac{W}{\eta}-W

Q_{L} = \left(\frac{1}{\eta}-1 \right)\cdot W(2)

Where:

Q_{H} - Heat received by the engine, measured in Btu.

Q_{L} - Heat rejected by the engine, measured in Btu.

W - Work output, measured in Btu.

If we know that \eta = 0.678 and Q_{L} = 55\,Btu, then the work output of the Carnot engine is:

W = \frac{Q_{L}}{\frac{1}{\eta}-1 }

W = 115.807\,Btu

The work output of the Carnot engine is 115.807 Btu. (Answer: C)

5 0
3 years ago
How will you charge a metal sphere by the process of induction?
zzz [600]
<span>A Conducting Sphere Is On Top Of An Insulating Stand. So A Negatively Charged Tube Is Brought Near The Neutral Sphere Then It Forces Electron Movement From The Left To The Right Side Of The Sphere. Once Touched By The Ground, The Electrons Leave The Sphere. When The Tube Is Moved Away, There's An Overall Positive Charge Left On The Sphere.</span>
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