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3 years ago

Which is the example of circular motion??

Physics

2 answers:

AnnyKZ [126]3 years ago

6 0

ferris whell

none of them

Diano4ka-milaya [45]3 years ago

3 0

Explanation:

<h2>1) Ferris Wheel</h2>

this is your answer

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What is the process by which water is transferred to the atmosphere by plants and trees called?

Ivan

Transpiration Is the process that happens when plants do this.

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3 years ago

Read 2 more answers

A ship leaves a port at noon and travels due west at 20 knots. At 6 PM, a second ship leaves the same port and travels northwest

Amanda [17]

Answer:

$v = 12.44 Knots$

Explanation:

First ship starts at Noon with speed 20 Knots towards West

now we know that 2nd ship starts at 6 PM with speed 15 Knots towards North West

so after time "t" of 2nd ship motion the two ships positions are given as

$r_1 = 20(t + 6)\hat i$

$r_2 = 15(t)(cos45\hat i + sin45\hat j)$

now we can find the distance between two ships as

$x = \sqrt{(20(t + 6) - 10.6 t)^2 + (10.6t)^2}$

now we have

$x^2 = (120 + 9.4 t)^2 + (10.6 t)^2$

$x^2 = 200.72 t^2 + 14400 + 2256 t$

now we will differentiate it with respect to time

$2x\frac{dx}{dt} = 401.44 t + 2256$

here we know that

$t = \frac{90}{15} = 6 hours$

so we have

$x = 187.5$

now we have

$2(187.5) v = 401.44(6) + 2256$

$v = 12.44 Knots$

5 0

3 years ago

George is planning an experiment to see how fast different sized cans of soup can roll down a ramp in his kitchen. He will be ro

ddd [48]

If it were me, I would choose D.

6 0

4 years ago

) A 100-g ball falls from a window that is 12 m above ground level and experiences no significant air resistance as it falls. Wh

yawa3891 [41]

Answer:

Momentum = 1.534 kgm/s

Explanation:

Using the equations of motion, we can obtain the velocity of the ball as it hits the ground.

g = 9.8 m/s²

y = 12 m

u = initial velocity = 0 m/s, since the ball was released from rest

v = final velocity befor the ball hits the ground.

v² = u² + 2ay

v² = 0 + 2×9.8×12 = 235.2

v = 15.34 m/s

The momentum at any point is given as mass × velocity at that point

Mass = 100 g = 0.1 kg, velocity = 15.34 m/s

Momentum = 0.1 × 15.34 = 1.534 kgm/s

3 0

4 years ago

A 8.0\,\text {kg}8.0kg8, point, 0, start text, k, g, end text box is released from rest at a height y_0 =0.25\,\text my 0 =0.2

Ratling [72]

Answer:

μ = 0.125

Explanation:

To solve this problem, which is generally asked for the coefficient of friction, we will use the conservation of energy.

Let's start working on the ramp

starting point. Highest point of the ramp

Em₀ = U = m h y

final point. Lower part of the ramp, before entering the rough surface

$Em_{f}$ = K = ½ m v²

as they indicate that there is no friction on the ramp

Em₀ = Em_{f}

m g y = ½ m v²

v = $\sqrt{2gy}$

we calculate

v = √(2 9.8 0.25)

v = 2.21 m / s

in the rough part we use the relationship between work and kinetic energy

W = ΔK = K_{f} -K₀

as it stops the final kinetic energy is zero

W = -K₀

The work is done by the friction force, which opposes the movement

W = - fr x

friction force has the expression

fr = μ N

let's write Newton's second law for the vertical axis

N-W = 0

N = W = m g

we substitute

-μ m g x = - ½ m v²

μ = $\frac{v^{2} }{2 g x}$

Let's calculate

μ = $\frac{2.21^{2}}{2\ 9.8\ 2.0}$

μ = 0.125

4 0

3 years ago