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3 years ago

7

A boy is swinging a yo-yo with mass 0.25 kg in a circle with radius 1 m at a speed of 6 m/s. What is the tension in the string c

onnecting the yo-yo to the boy's finger, disregarding the effect of gravity?
A. 20N
B. 12N
C. 35N
D. 9N

1 answer:

Paladinen [302]3 years ago

5 0

Answer:

D. 9 N

Explanation:

The tension on the string is equivalent to the centripetal force.

Centripetal force is the force exerted by an object in circular motion or path towards the center of the circular path.

Centripetal force = mv²/r

where m is the mass of the object, v is the velocity and r is the radius of the circular path.

Centripetal force = (0.25 kg × 6²)/ 1

= 9 N

Thus, the tension on the string is 9 N

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The velocity of the transverse waves produced by an earthquake is 5.08 km/s, while that of the longitudinal waves is 8.3312 km/s

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Answer:

727.67 km

Explanation:

Sine they have Same distance D

distance = speed * time

D = 5.08t

D = 8.3312(t+55.9)

so

5.08t = 8.3312(t+55.9) t in

3.2512t = 465.71

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Subtitute t

D=5.08 t

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Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5

LekaFEV [45]

Answer:

$|\vec{F}_3| = 102.92 \ N$

$\theta = 57 \° 24 ' 48''$

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

$\vec{a} = \vec{0}$ .

As the net force equals acceleration multiplied by mass , this must mean:

$\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}$ .

So, the sum of the three forces must be zero:

$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$ ,

this implies:

$\vec{F}_3 = - \vec{F}_1 - \vec{F}_2$ .

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector $\hat{i}$ pointing east, with the y axis unit vector $\hat{j}$ pointing south, so the positive angle is south of east. For this, we got for the first force:

$\vec{F}_1 = 83.7 \ N \ (-\hat{j})$ ,

as is pointing north, and for the second force:

$\vec{F}_2 = 59.9 \ N \ (-\hat{i})$ ,

as is pointing west.

Now, our third force will be:

$\vec{F}_3 = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})$

$\vec{F}_3 = 83.7 \ N \ \hat{j} + 59.9 \ N \ \hat{i}$

$\vec{F}_3 = (59.9 \ N , 83.7 \ N )$

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}$

$|\vec{F}_3| = 102.92 \ N$

this is the magnitude.

To find the direction, we can use:

$\theta = arctan(\frac{F_{3_y}}{F_{3_x}})$

$\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })$

$\theta = 57 \° 24 ' 48''$

and this is the angle south of east.

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The ball is displaced to the left and then oscillates backwards and forwards between the two plates. The ball touches a plate on

ELEN [110]

Answer:

Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere

Explanation:

The formula to be used here is

Q = It

where Q is the quantity of electricity and it is measured coulombs (C); 2.8 × 10⁻⁸ C or 0.000000028 C

I is current and it is measured in ampere (amps or A); unknown

t is time and it is measured in seconds (s); 0.05 s

Since, average current is what is unknown

I =Q/t

I = 0.000000028/0.05

I = 5.6 × 10⁻⁷ A

Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere

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A simple electric motor consists of a 220-turn coil, 4.4 cm in diameter, mounted between the poles of a magnet that produces a 9

Verdich [7]

Answer:

$M = 5.01\ A.m^2$

Explanation:

It is given that,

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Diameter of the coil, d = 4.4 cm

Radius of the coil, r = 2.2 cm = 0.022 m

Magnetic field produced by the poles of magnet, $B=96\ mT=96\times 10^{-3}\ T$

Current flowing in the coil, I = 15 A

Let M is the coil's magnetic dipole moment. Its formula is given by :

$M=N\times I\times A$

$M=220\times 15\times \pi (0.022)^2$

$M = 5.01\ A.m^2$

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