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3 years ago

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a kid on a playground swing makes 6 complete to-and-from swings each 30 seconds (a) the frequency of the swinging is? (b) the pe

riod of the swinging is ?

2 answers:

Scorpion4ik [409]3 years ago

6 0

Show Me The Money 9 will likely be postponed until the fall instead of the summer, similar to the air dates of Show Me The Money 777, because of the Kung Flu pandemic going around.

Mariana [72]3 years ago

4 0

The frequency is how many per second:

(6 swings)/(30 sec) = (6/30) swing/sec = 0.2 per sec = 0.2 Hz .

The period is how long each one takes, or seconds per swing.
It's exactly the flip of frequency.
So we could just take the frequency, flip it, and find 1 / 0.2 ,
but let's do it the long way:

(30 sec) / (6 swings) = (30/6) sec/swing = 5 seconds .

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g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

2 years ago

A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude

Darya [45]

Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

$\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s$

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

8 0

2 years ago

Calculate the speed of a car that traveled 200 kilometers in 3 hours.

nignag [31]

Answer:

The units (km/h) tell you how to do this! 200km/3h = 66.66666666…. BUT technically you only have ONE significant digit: 3 so 66.666… rounded to ONE digit is 70km/h but that is probably not important in this intro class so V = 66.67 or 67 km/h

8 0

2 years ago

Read 2 more answers

a person was using a spanner to loosen a tight nut.Their mass was 50kg the spanner was 0.5m long i) what is the persons weight?

sineoko [7]

1) Weight of the person: 490 N

2) Maximum torque: 245 Nm

Explanation:

1)

The weight of a body is equal to the gravitational force exerted on the body; it is given by the equation

$F=mg$

where

m is the mass of the body

g is the acceleration due to gravity

For the person in this problem,

m = 50 kg is the mass

$g=9.8 m/s^2$ is the acceleration due to gravity on Earth's surface

Therefore, the weight of the person is

$F=(50)(9.8)=490 N$

2)

The turning force (also called torque) exerted by a force rotating an object is given by

$\tau = Fd sin \theta$

where

F is the magnitude of the force

d is the length of the arm (the distance between the force and the pivotal point)

$\theta$ is the angle between the direction of the force and the arm

For the spanner in this problem,

F = 490 N is the force applied (the weight of the person)

d = 0.5 m is the arm (the length of the spanner)

The maximum torque is obtained when $\theta=90^{\circ}$ , therefore it is:

$\tau=(490)(0.5)(sin 90^{\circ})=245 N\cdot m$

Learn more about weight:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

About torque:

brainly.com/question/5352966

#LearnwithBrainly

5 0

3 years ago

Failures to arrive at an answer in a scientific investigation can still be useful because they show us what doesn't work. Answer

kipiarov [429]

True
From what Ive learned its like how you create your own hypothesis for a science experiment it may not work completely that way but you learn what didnt work and what does work

4 0

3 years ago