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Triss [41]
2 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]2 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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Alyssa is carrying a water balloon while running down a field at a speed of 14 m/s. She tosses the water balloon forward toward
Luda [366]
From Alyssa's point of view, the water balloon is at first at rest and then gets thrown with a velocity of 23m/s. Therefore the balloon will have a speed of 23m/s for Alyssa.

At the same time, Naya is watching, and she sees the balloon at the beginning moving at a speed of 14m/s along with Alyssa, and then pushed forward of other 23m/s. Therefore, from her point of view, the balloon will have a speed of 14+23 = 37m/s.

Hence, the correct answer is <span>D) The speed of the balloon is 23 m/s for Alyssa and 37 m/s for Naya. </span>
4 0
2 years ago
Read 2 more answers
The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the ob
givi [52]

Answer:

Charge, Q=1.56\times 10^{-8}\ C

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

E=\dfrac{kQ}{r^2}

Q is the charge on an object

Q=\dfrac{Er^2}{k}

Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}

Q=1.56\times 10^{-8}\ C

So, the charge on the object is 1.56\times 10^{-8}\ C. Hence, this is the required solution.

7 0
2 years ago
A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f
konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m

(c) the maximum speed attained by the object during its motion

E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s

8 0
2 years ago
The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vert
zimovet [89]

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

F=\dfrac{mv^2}{r}

v is the maximum speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s

Hence, the maximum speed of the ball is 5.86 m/s.

3 0
2 years ago
You start with spring that's already been stretched an unknown amount from equilibrium. After stretching it an additional 2.0 cm
maxonik [38]

Answer: 35*10^3 N/m

Explanation: In order to explain this problem we know that the potential energy for spring is given by:

Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.

We  also know that with additional streching of 2 cm of teh spring,  the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

k=2*7/(0.02)^2=35000 N/m

7 0
2 years ago
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