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3 years ago

May someone please help?

Physics

1 answer:

attashe74 [19]3 years ago

6 0

Answer:

CONDUCTION

Energy transfers from one

object to another by direct

contact

CONVECTION

Energy transfers via the

movement of particles

within a liquid or gas.

RADIATION

Energy transfers by

electromagnetic waves as

they travel through space or

matter

Explanation:

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What is imperceptible human motion?

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Unable to be noticed or not felt because of feeling slight

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3 years ago

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

Gemiola [76]

Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

$v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}$

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}$

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}$

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

$\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m$

THe compression of the ball when it strikes the floor is 5.11*10^-3m

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4 years ago

What force holds the earth-moon system together?

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Gravity holds the system together

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3 years ago

Read 2 more answers

Bus starts from rest if the acceleration of the bus is 0.5 MS squared what will be the velocity at the end of two minutes and wh

Nutka1998 [239]

Explanation:

Given that,

Initial speed of the bus, u = 0

Acceleration of the bus, a = 0.5 m/s²

Let v is the velocity at the end of 2 minutes. The change in velocity divided by time equals acceleration.

So,

$a=\dfrac{v-u}{t}\\\\v=u+at\\\\v=0+0.5\times 120\\\\v=60\ m/s$

Let d is the distance cover during that time. So,

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(60)^2}{2\times 0.5}\\\\d=3600\ m$

So, the final speed is 60 m/s and the distance covered during that time is 3600 m.

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3 years ago

Anyone knows this? Please answer... Spam will be reported.

Yakvenalex [24]

Answer:

The correct option is;

The assertion is correct, but reason wrong

Explanation:

The question is with regards to the relationship between work, energy, power, and velocity

The mass of each of the persons running up the staircase = Different

The time it takes each person to run up the stairs = Equal time

Let, 'm₁' and 'm₂' represent the mass of each of the persons that ran up the stairs and m₁ > m₂

Let 't' represent the equal time it takes then to run up the stairs

Let 'h' represent the height of the stairs

The energy, 'E', it takes to run up the stairs is equal to the potential energy, P.E., obtained at the top of the stairs

P.E. = m·g·h

Where;

m = The mass of the person at an elevated height

g = The acceleration due to gravity = Constant

h = The height reached above ground level

Given that the height reached is the same for both of the persons, we have

For m₁, P.E.₁ = m₁·g·h and for m₂, P.E.₂ = m₂·g·h

Therefore, where, m₁ > m₂, we have;

P.E.₁ > P.E.₂

∴ E₁ > E₂

Power, 'P', is the rate at which energy is expended

∴ Power, P = E/t

∴ P₁ = E₁/t > P₂ = E₂/t

Therefore, the person with the greater mass, 'm₁', uses more power than the person of mass 'm₂', in running up the stairs

Therefore, the assertion is correct

The average velocity, vₐ = (Total distance traveled, d)/(Total time taken, t)

Given that the distance, 'd', covered in running up the stairs by both persons is the same, and the time it takes them to complete the distance, 't', is also the same, we have;

The average velocity of the person with the greater mass m₁ is the same as the average velocity of the person with mass, m₂

Therefore, the reason is wrong

The answer is that the assertion is correct, but reason wrong

6 0

3 years ago