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4 years ago

Solar System Model Comparison and Contrast Chart

Physics

1 answer:

kotykmax [81]4 years ago

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A 10n force is applied to a 25kg mass to slide it across a frictional surface. What is the acceleration of the mass?

klasskru [66]

Answer: a = 0.4m/s^2 - 9.8*c where c is the coefficient of kinetic friction of the surface

Explanation: We know that, by the second Newton's law, a = F/m

where a is the acceleration, F is the net force and m is the mass of the object.

Then, if the surface is frictionless, the total force applied in the object is 10N, and the mass of the object is 25kg, so the acceleration is:

a =10N/25kg = 0.4m/s^2.

But if the surface is frictional, there will be a force of friction applied in the mass (this depends on the coefficient of friction and the weight of the mass), this means that the acceleration will be reduced.

If = -(9.8*25)*c

where c is a number that is bigger than 0 and smaller than 1, is called the coefficient of kinetic friction.

So the total force is now:

F = (10 - 9.8*25*c)

Then, the acceleration in a frictional surface is equal to:

a = (10 - 9.8*25*c)/25 = 0.4m/s^2 - 9.8*c

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4 years ago

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PLZ Help Me I give you brainlist also NO LONKS OR I REPORT and Have a gr8 day my peeps

Sergio039 [100]

Answer:

the answer is a because I saw it in a syllabus

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3 years ago

420 hg = _____ cg help please

kondaur [170]

4200000 is your answer hope this helps

4 0

3 years ago

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What is the wavelength of a 6.00*10^2 Hz sound wave in air at 20C

Alex Ar [27]

Answer:

Solution

λ=v/n

Here, v=344 m s−1

n=22 MHz =22×106 Hz

λ=344/22×106=15.64×10−6m=15.64μm.

5 0

2 years ago

Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of

morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner $(v_1)$ =3.1 m/s

speed of second runner $(v_2)$ =4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner $t=\frac{295.6}{4.65}=63.56 s$

time required by first runner to reach finish line $=\frac{250}{3.1}=80.64 s$

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

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4 years ago