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3 years ago

A. A child is twirling a 1.52 kg object in a vertical circle with a radius of 67.6

Physics

1 answer:

steposvetlana [31]3 years ago

6 0

Answer:

(a) 4.21 m/s

(b) 24.9 N

Explanation:

(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.

Sum the forces in the radial (+y) direction:

∑F = ma

T − mg = m v² / r

v = √(r (T − mg) / m)

v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)

v = 4.21 m/s

(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.

Sum the forces in the radial (-y) direction:

∑F = ma

T + mg = m v² / r

T = m v² / r − mg

T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)

T = 24.9 N

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2 years ago

the minimum flying speee for a bird called a house martin is 9.0ms^-1. it reaches this speed by falling from its nest before swo

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Answer:

The nest must be about 4.15 meters above ground

Explanation:

Use the velocity equation under accelerated motion (acceleration of gravity ):

$v_f=v_i+a\,*\,t$

which for this case has initial velocity = 0 (falls from the nest), final velocity = 9 m/s, and a = 9.8 m/s^2, then we can find the time needed in air while falling to reach the required speed:

$v_f=v_i+a\,*\,t\\9=0+9.8\,t\\t=\frac{9}{9.8} \, sec\\t \approx 0.92\,\, sec$

We now use this time value to find the distance covered in free fall during 0.92 seconds:

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2 years ago

A constant magnetic field passes through a single rectangular loop whose dimensions are 0.42 m x 0.54 m. The magnetic field has

olga_2 [115]

Answer:

$0.29887\ \text{V}$

Explanation:

A = Area = $0.42\times0.54\ \text{m}^2$

B = Magnetic field = 1.7 T

$\theta$ = Angle that magnetic field makes with the plane of the loop = $71^{\circ}$

t = Time = 0.42 s

EMF is given by

$\varepsilon=A\cos\theta\dfrac{dB}{dt}\\ =\dfrac{0.42\times 0.54\times\cos 71^{\circ}\times 1.7}{0.42}\\ =0.29887\ \text{V}$

The magnitude of the average emf induced in the loop is $0.29887\ \text{V}$ .

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2 years ago

Describe how different types of motion are represented by distance-time graphs and velocity-time graphs.

eduard

Answer:

non-accelerated movement

velocity versus time a horizontal straight line.

distance versus time gives a horizontal straight line.

accelerated motion

graph of velocity versus time s an inclined line and the slope

graph of distance versus time is a parabola of the form

Explanation:

In kinematics there are two types of steely and non-accelerated movements

In a the velocity of the body is constant therefore a speed hook against time gives a horizontal straight line.

A graph of distance versus time is a straight line whose slope is the velocity of the body

x = v t

In an accelerated motion the velocity changes linearly with time, so a graph of velocity versus time is an inclined line and the slope is the value of the acceleration of the body

v = v₀ + a t

A graph of distance versus time is a parabola of the form

x =v₀ t + ½ a t²

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2 years ago