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katrin2010 [14]
3 years ago
12

During a trial run, race car A starts from rest and accelerates uniformly along a straight level track for a particular interval

of time. Race car B also starts from rest and accelerates at the same rate, but for twice the time. At the end of their respective acceleration periods, which of the following statements is true? Car A has traveled a greater distance. Car B has traveled twice as far as A. Car B has traveled four times as far as car A. Both cars have traveled the same distance
Physics
1 answer:
kondor19780726 [428]3 years ago
5 0
Answer: car B has travelled 4times as far as Car A

d=vi*t+1/2at^2

No initial velocity so equation becomes;

d=1/2at^2 and the acceleration is the same between both only time is different;

Car A d=1/2a(1)^2

Car B d=1/2a(2)^2

Car A d= 1^2=1

Car B d= 2^2=4

Car B d=4*Car A

So car B has travelled 4 times as far as car A
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Sirius A has a luminosity of 26 LSun and a surface temperature of about 9400 K.What is its radius?
wlad13 [49]

Answer

Given,

Sirius A surface temperature,T = 9400 K

Sirius A luminosity,L = 26 L₀

L₀ is the luminosity of sun.

Radius of sun =  695700000 m

Temperature on sun surface = 5780 K

Luminous intensity is given by:-

L=4 \pi R^{2} \sigma T^{4}

Now

\frac{L}{L_{0}}=\frac{4 \pi R^{2} \sigma T^{4}}{4 \pi R_{0}^{2} \sigma T_{0}^{4}}=26

\Rightarrow \frac{R^{2} T^{4}}{R_{0}^{2} T_{0}^{4}}=26

\Rightarrow R^{2}=26 \times \frac{R^{2} T_{0}^{4}}{T^{4}}=26 \times \frac{695700000^{2} \times 5780^{4}}{9400^{4}}

R=1341246640\ m=1.34 \times 10^{9}

7 0
3 years ago
An object weighs 60.0 kg on the surface of the earth. How much does it weigh 4R from the surface? (5R from the center)
Alecsey [184]
"60 kg" is not a weight.  It's a mass, and it's always the same
no matter where the object goes.

The weight of the object is   

                                 (mass) x (gravity in the place where the object is) .

On the surface of the Earth,

                   Weight = (60 kg) x (9.8 m/s²)

                                =      588 Newtons.

Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to  5R  from the center, the gravity out there is

                    (1R/5R)²  =  (1/5)²  =  1/25  =  0.04 of its value on the surface.

The object's weight would also be 0.04 of its weight on the surface.

                 (0.04) x (588 Newtons)  =  23.52 Newtons.

Again, the object's mass is still 60 kg out there.
___________________________________________

If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink. 
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.
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What is the difference between a spring and a stream?
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evablogger [386]

Answer:

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