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Triss [41]
4 years ago
7

After getting a haircut, Joey’s barber spins him around in his barber’s chair 2 times per second. Is period or frequency given?

________________ What is the period? ______ What is the frequency? __________ (if you can please answer the other questions, would really appreciate it)

Physics
1 answer:
Rzqust [24]4 years ago
7 0

Answer:

Explanation:

This figure given is the frequency; 2 times per second represents frequency.

What is frequency?

  • It is the number of times per seconds something goes past or around another.

 it is expressed as:

            Frequency  = \frac{n}{t}

                  where n is the number of turns

                              t is the time taken

  Therefore, the Barber spinned him 2 times in 1 second.

The period is the inverse of frequency. It is the time taken for a body to go through a point;

              Period  = \frac{t}{n }   = \frac{1}{f}    = \frac{1}{2}s

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A catcher in a baseball game stops a pitched ball that was originally moving at 44 m/s over a distance of 12.5 cm. the mass of t
eduard

Answer:

Force = -1161.6 Newton

Explanation:

Given the following data;

Initial velocity, u = 44m/s

Distance ,s = 12.5cm to m = 12.5/100 = 0.125m

Mass = 0.15kg

To find the acceleration;

We would use the third equation of motion;

V ² = U² + 2as

0² = 44² + 2*a*0.125

0 = 1936 + 0.25a

0.25a = -1936

a = -1936/0.25

Acceleration, a = -7744m/s2

Force = mass * acceleration

Substituting into the equation, we have;

Force = 0.15 * (-7744)

Force = -1161.6 Newton

The value of its force is negative because the glove decreases the velocity of the ball.

4 0
3 years ago
What is the term for the overall change in position?
mart [117]
That's "displacement". It only depends on the beginning and ending locations, and doesn't care about the route between them.
6 0
3 years ago
Read 2 more answers
Bill throws a tennis ball to his dog. He throws the ball at a speed of 15 m/s at an angle of 30° to the horizontal. Assume he th
Sidana [21]

1a) Bill and the dog must have a speed of 13.0 m/s

1b) The speed of the dog must be 22.5 m/s

2a) The ball passes over the outfielder's head at 3.33 s

2b) The ball passes 1.2 m above the glove

2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit

2d) One solution is when the player is jumping up, the other solution is when the player is falling down

Explanation:

1a)

The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.

The ball's initial speed is

u = 15 m/s

And the angle of projection is

\theta=30^{\circ}

So, the ball's horizontal velocity is

v_x = u cos \theta = (15)(cos 30)=13.0 m/s

And therefore, Bill and the dog must have this speed.

1b)

For this part, we have to consider the vertical motion of the ball first.

The vertical position of the ball at time t is given by

y=u_yt+\frac{1}{2}at^2

where

u_y = u sin \theta = (15)(sin 30) = 7.5 m/s is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity

The ball is at a position of y = 2 m above the ground when:

2=7.5t + \frac{1}{2}(-9.8)t^2\\4.9t^2-7.5t+2=0

Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.

The horizontal distance covered by the ball during this time is

d=v_x t =(13.0)(1.19)=15.5 m

The dog must be there 0.5 s before, so at a time

t' = t - 0.5 = 0.69 s

So, the speed of the dog must be

v_x' = \frac{d}{t'}=\frac{15.5}{0.69}=22.5 m/s

2a)

Here we just need to consider the horizontal motion of the ball.

The horizontal distance covered is

d=98 m

while the horizontal velocity of the ball is

v_x = u cos \theta = (34)(cos 30)=29.4 m/s

where u = 34 m/s is the initial speed.

So, the time taken for the ball to cover this distance is

t=\frac{d}{v_x}=\frac{98}{29.4}=3.33 s

2b)

Here we need to calculate the vertical position of the ball at t = 3.33 s.

The vertical position is given by

y= h + u_y t + \frac{1}{2}at^2

where

h = 1.2 m is the initial height

u_y = u sin \theta = (34)(sin 30)=17.0 m/s is the initial vertical velocity

a=g=-9.8 m/s^2 is the acceleration of gravity

Substituting t = 3.33 s,

y=1.2+(17)(3.33)+\frac{1}{2}(-9.8)(3.33)^2=3.5 m

And sinc the glove is at a height of y' = 2.3 m, the difference in height is

y - y' = 3.5 - 2.3 = 1.2 m

2c)

In order to intercept the ball, he jumps upward at a vertical speed of

u_y' = 7 m/s

So its position of the glove at time t' is

y'= h' + u_y' t' + \frac{1}{2}at'^2

where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be

y' = y = 3.5 m (the height of the ball)

Substituting and solving for t', we find

3.5 = 2.3 + 7t' -4.9t'^2\\4.9t'^2-7t'+12 = 0

Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is

t'' = t -t'

So we have two solutions:

t'' = 3.33 s - 0.20 s = 3.13 s\\t'' = 3.33 s - 1.23 s = 2.10 s

So, the player can jump after 2.10 s or after 3.13 s.

2d)

The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he  falls down.

Therefore, the two solutions corresponds to the two different part of the motion.

The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.

On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0
3 years ago
For the following elementary reaction 2br• -> br2-. The rate of consumption of the reaction and the rate of formation of prod
Scorpion4ik [409]

Answer: -\frac{1}{2}\times \frac{d[Br^.]}{dt}=+\frac{d[Br_2]}{dt}

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

2Br^.\rightarrow Br_2

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{d[Br^.]}{2dt}

or Rate=+\frac{d[Br_2]}{dt}

Thus -\frac{d[Br^.]}{2dt}=+\frac{d[Br_2]}{dt}

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Answer:

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Explanation:

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