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3 years ago

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A constant force of magnitude F acts on an object of mass 0.04kg initially at rest at a point O. If the speed of the object when

it has moved 50m from O is 500m/s. what is the value of F?

1 answer:

vampirchik [111]3 years ago

5 0

Answer:

F = 100 Newtons

Explanation:

F = ?

m = 0.04kg

u = 0m/s ==> u is just an abbreviation for initial velocity, it is conventional.

s = 50m ==> s is just an abbreviation for distance, it is conventional.

v = 500m/s ==> v is just an abbreviation for final velocity, it is conventional.

$v^{2} = u^{2} + 2as\\\\=> a = \frac{v^{2} - u^{2}}{2s}\\a = \frac{500^{2} }{2*50}\\a = 2500ms^{-2}$

Then F = ma = 0.04 x 2500 = 100N

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Two cylindrical rods, one copper and the other iron, are identical in lengths and cross-sectional areas. They are joined, end to

Pie

Answer:

Vc = 2.41 v

Explanation:

voltage (v) = 16 v

find the voltage between the ends of the copper rods .

applying the voltage divider theorem

Vc = V x ( $\frac{Rc}{Rc + Ri}$ )

where

Rc = resistance of copper = $\frac{ρl}{a}$ (l = length , a = area, ρ = resistivity of copper)

Ri = resistance of iron = $\frac{ρ₀l}{a}$ (l = length , a = area, ρ₀ = resistivity of copper)

Vc = V x ( $\frac{\frac{ρl}{a}}{\frac{ρl}{a} + \frac{ρ₀l}{a}}$ )

Vc = V x ( $\frac{ρ x (\frac{l}{a})}{(ρ + ρ₀) x (\frac{l}{a})}$ )

Vc = V x ( $\frac{ρ}{ρ + ρ₀}$ )

where

ρ = resistivity of copper = 1.72 x 10^{-8} ohm.meter

ρ₀ = resistivity of iron = 9.71 x 10^{-8} ohm.meter

Vc = 16 x ( $\frac{1.72 x 10^{-8}}{1.72 x 10^{-8} + 9.71 x 10^{-8}}$ )

Vc = 2.41 v

5 0

3 years ago

A man starts walking north at 3 ft/s from a point P. Five minutes later a woman starts walking south at 4 ft/s from a point 500

mrs_skeptik [129]

Answer:

ds/dt = 6.98 ft/s

Explanation:

Given:

- The speed of man due north Vm = 3 ft/s

- The speed of woman due south Vw = 4 ft/s

- Woman starts walking 5 mins later than man

Find:

At what rate are the people moving apart 15 min after the woman starts walking?

Solution:

- The total time for which the man is walking due north from P, is Tm:

Tm = 5 + 15 = 20 mins

- The total distance traveled by man in Tm mins is:

Dm = Tm*Vm

Dm = 20*60*3

Dm = 3,600 ft

- The total time for which the woman is walking due south from 500 ft due east from P, is Tw:

Tw = 15 = 15 mins

- The total distance traveled by man in Tw mins is:

Dw = Tw*Vw

Dw = 15*60*4

Dw = 3,600 ft

- The displacement between man and woman at any instance is (s) which can be related by pythagoras theorem as follows:

s^2 = (dm + dw)^2 + 500^2

Where, dm : Distance travelled by man at any time Tm

dw : Distance travelled by woman at any time Tw

- Differentiate s with respect to t:

2s*ds/dt = 2*(dm + dw)*(Vm + Vw)

s*ds/dt = (dm + dw)*(Vm + Vw)

ds/dt = [ (dm + dw)*(Vm + Vw) ] / s

- Evaluate the rate of separation of man and woman ds/dt by evaluating at instance Tm = 20 mins and Tw = 15 mins. We have:

ds/dt = [ (Dm + Dw)*(Vm + Vw) ] / sqrt ( (Dm + Dw)^2 + 500^2 )

- Plug in the values:

ds/dt = [ (3600 + 3600)*(3 + 4) ] / [sqrt ( (3600 + 3600)^2 + 500^2 )]

ds/dt = 6.98 ft/s

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3 years ago

Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back an

disa [49]

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = $\frac{1}{2} *m*vf^{2}$ (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

$vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s$

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

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3 years ago

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Liono4ka [1.6K]

The correct answer is D. Newton

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3 years ago

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In a pool game, the cue ball, which has an initial speed of 3.0 m/s, make an elastic collision with the eight ball, which is ini

zhenek [66]

Explanation:

Given

initial speed(u)=3 m/s

mass of each ball is m

Let the cue ball is moving in x direction initially

In elastic collision Energy and momentum is conserved

Let u be the initial velocity and $v_1 , v_2$ be the final velocity of 8 ball and cue ball respectively

$\frac{mu^2}{2}+0=\frac{mv^2_1}{2}+\frac{mv^2_2}{2}$

The angle after which cue ball is deflected is given by

$\theta _1=90-40=50^{\circ}$

Conserving momentum in x direction

$mu=mv_1cos40+mv_2cos50$

$3=v_1cos40+v_2cos50$

Along Y axis

$0+0=v_1sin40-v_2sin50$

$v_1sin40=v_2sin50$

substitute the value of $v_1$

we get $v_2=1.912 m/s$

$v_1=2.27 m/s$

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3 years ago