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2 years ago

6

Tapping the surface of a pan of water generates 17.5 waves per second. If the wavelength of each wave is 45 cm, what is the spee

d of the waves?

1 answer:

ElenaW [278]2 years ago

5 0

Answer:

Speed of the wave is 7.87 m/s.

Explanation:

It is given that, tapping the surface of a pan of water generates 17.5 waves per second.

We know that the number of waves per second is called the frequency of a wave.

So, f = 17.5 Hz

Wavelength of each wave, $\lambda=45\ cm=0.45\ m$

Speed of the wave is given by :

$v=f\lambda$

$v=17.5\times 0.45$

v = 7.87 m/s

So, the speed of the wave is 7.87 m/s. Hence, this is the required solution.

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A bicycle tire rotates 25 times in 10 seconds. What is it’s average angular velocity?

ryzh [129]

<h2>Answer:</h2>

<u>Angular velocity of bicycle tire is 15.78 radians per second.</u>

<h3>Explanation:</h3>

Angular velocity is the change in angular speed of an object with respect to time take for change or it is the rate of change of circular motion.

In the given question the circular displacement is 25 rounds around a central point.

The angular displacement is measured in degrees and 1 round is equal to 360 degrees.

25 Rounds = 25 × 360 = 9000 degrees.

Angular velocity = angular displacement /time = 9000/10 = 900 degrees per second.

In SI,angular velocity is represented in radians per second.

So, 1 radian = 57.29 degrees

Angular velocity = 15.78 radians per second

3 0

2 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

2 years ago

A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in t

stiv31 [10]

Answer:

D) 117 rad/s

Explanation:

We can treat this system as a circular motion where the origin is the elbow joint, the ball rotation velocity v is 35 m/s, the rotation radius is r = 0.3m.

As the ball is leaving the pitcher hand at such speed and rotation radius. Its angular velocity is:

$\omega = \frac{v}{r} = \frac{35}{0.3} = 117 rad/s$

3 0

2 years ago

Cori uses 475 J of energy from her muscles to push a bar 1 m on a weight machine at the gym. Between the bar’s motion , the heat

dimulka [17.4K]

Answer:

475

Explanation:

Cori does not exert any more force than 475 J, so 475 is the answer.

5 0

2 years ago

Read 2 more answers

a child hits a ball with a force of 350 N. (a) If the ball and bat are in contact for 0.12 is, what impulse does the ball receiv

Lina20 [59]

Explanation:

Given that,

Force with which a child hits a ball is 350 N

Time of contact is 0.12 s

We need to find the impulse received by the ball. The impulse delivered is given by :

$J=F\times t\\\\J=350\times 0.12\\\\J=42\ N-m$

So, the impulse is 42 N-m..

We know that he change in momentum is also equal to the impulse delivered.

So, impulse = 42 N-m and change in momentum =42 N-m.

7 0

2 years ago