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2 years ago

If a man does 4500 J of work from a high rise building of 10 m, what is his mass?

Physics

2 answers:

GalinKa [24]2 years ago

4 0

Explanation:

The potential energy (Ep) = 4500 J

Ep = m. g. h

=> m = Ep/(g.h)

m = 4500/(10×10)

= 4500/100

= 45

so, the mass = 45 kgs

san4es73 [151]2 years ago

3 0

Im not sure 275928237491

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How does light travel across the universe to earth?

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Light travels at the speed of 186,000 miles per second. If you were to travel around the earth it would be 7.5 times in a second

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2 years ago

A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

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3 years ago

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eduard

Answer:

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2 years ago

A 62 pound box is on an incline. Determine the minimum force P that would result in the box starting to slide up the incline. (i

Anon25 [30]

Answer:

F > W * sin(α)

Explanation:

The force needed for the box to start sliding up depends on the incline (α).

The external forces acting on the box would be the weight, the normal reaction and the lifting force that is applied to make it slide up.

These forces can be decomposed on their normal and tangential (to the slide plane) components.

The weight will be split into

Wn = W * cos(α) (in normal direction)

Wt = W * sin(α) (in tangential direction)

The normal reaction will be alligned with the normal axis, and will be equal to -Wn

N = -W* cos(α) (in normal direction)

To mke the box slide up, a force must be applied, that is opposite to the tangential component of the weight and at least a little larger

F > |-W * sin(α)| (in tangential direction)

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3 years ago

What heat transfer are uv rays from the sun. Radiation, conduction, or convection

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Answer:

The ans is Radiation

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