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2 years ago

9

A 22.0-kg child is riding a playground merrygo-round that is rotating at 40.0 rev/min. What centripetal force is exerted if he i

s 1.25 m from its center?

1 answer:

Tems11 [23]2 years ago

6 0

Answer:

F=480.491 N

Explanation:

Given that

mass ,m = 22 kg

Angular speed ω = 40 rev/min

$\omega=\dfrac{2\pi \times 40}{60}\ rad/s$

ω =4.18 rad/s

The radius r= 1.25 m

We know that centripetal force is given as

F=m ω² r

Now by putting the values in the above equation we get

$F=22\times 4.18^2\times 1.25\ N$

F=480.491 N

Therefore the centripetal force on the child will be 480.491 N.

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Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

E = 3.84*10^{-4} N/C

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3 years ago

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Which describes how to generate a real image of an object using a concave mirror?The object must be closer to the mirror than th

iren2701 [21]

In order to form a real image using a concave mirror, the object must be placed beyond the center of curvature of the mirror. Therefore, the object must be further from the mirror than the focal point. The image that will form will be real, but it will also be inverted and its magnification will be less than 1, meaning it will be smaller than the actual object.

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2 years ago

What do you mean by population size explain

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Population size is the actual number of individuals in a population. Population density is a measurement of population size per unit area, i.e., population size divided by total land area. Abundance refers to the relative representation of a species in a particular ecosystem.

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3 years ago

A dog, with a mass of 10.0 kg, is standing on a flatboat so that he is 22.5 m from the shore. He walks 7.8 m on the boat toward

marissa [1.9K]

Answer:16.096

Explanation:

Given

mass of dog $\left ( m_d\right )=10kg$

mass of boat $\left ( m_b\right )=46kg$

distance moved by dog relative to ground= $x_d$

distance moved by boat relative to ground= $x_b$

Distance moved by dog relative to boat=7.8m

There no net force on the system therefore centre of mass of system remains at its position

0= $m_d\times x_d+m_b\dot x_b$

0= $10\times x_d+46\dot x_b$

$x_d=-4.6x_b$

i.e. boat will move opposite to the direction of dog

Now

$|x_d|+|x_b|=7.8$

substituting $x_d$ value

$5.6|x_b|=7.8$

$|x_b|=1.392m$

$|x_d|=6.4032m$

now the dog is 22.5-6.403=16.096m from shore

4 0

2 years ago

Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the mete

disa [49]

Answer:

Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.

(a) Determine the position of wire 3.

b) Determine the magnitude and direction of current in wire 3

Explanation:

a) $F_{net} \text {on wire }3=0$

$\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm$

position of wire = 50 - 1.2

= 48.8cm

b) $F_{net} \text {on wire }1=0$

$\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A$

Direction ⇒ downward

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3 years ago