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GalinKa [24]
4 years ago
7

In the oscillating spring ball system, what is true about the energy of the ball when it is located at its amplitude? Select the

correct answer A. The kinetic energy is at its maximum value and the potential energy is zero. B. The potential energy is at its maximum value and the kinetic energy is zero. C. The total energy is zero. D. Both the kinetic and potential energy are at their maximum our swer values
Physics
1 answer:
Nitella [24]4 years ago
8 0

When the maximum amplitude is reached, the body is also positioned in the maximum position which it can reach. At that point, the maximum potential energy stored in the system will therefore be taken.

PE = \frac{1}{2} kh_{max} ^2

However, when it reaches the maximum amplitude it is also in a change of direction of the velocity vector, reaching its minimum point in magnitude, thereby generating the kinetic energy to become zero.

KE = \frac{1}{2} mv^2

Therefore of the options shown, the correct answer would be B.

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What is the name of the theory describing how the lithosphere is broken into segments, or plates, which "float" on the asthenosp
kogti [31]

Answer:

C. Plate Tectonics

Explanation:

The theory of plate tectonics is when the lithosphere is separated into plates. These plates move over or float over the asthenosphere. The movement of these plates cause earthquakes and can interact with the volcanic activity.

4 0
3 years ago
8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott
nasty-shy [4]

<u>Answer:</u> The ball is travelling with a speed of 5.5 m/s after hitting the <u>bottle.</u>

<u>Explanation:</u>

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m_1,u_1\text{ and }v_1 are the mass, initial velocity and final velocity of ball.

m_2,u_2\text{ and }v_2 are the mass, initial velocity and final velocity of bottle.

We are given:

m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s

Putting values in above equation, we get:

(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

5 0
3 years ago
Why water in earthern pot remain cool in summer
hjlf

Answer:

Since in summer, the eastern side do not face the sunlight and hence the water in eastern pot remain cool in summer.

8 0
3 years ago
Read 2 more answers
At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th
Ainat [17]

Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

 ac is 6.23*g which is 9.81

v is 101m/s

substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r  

So (86kg* 101^2)/(167) =

Fc=5253 N

4 0
4 years ago
An isotropic point source emits light at wavelength 510 nm, at the rate of 170 W. A light detector is positioned 410 m from the
Wewaii [24]

Answer:

\frac{dB}{dt} = 3.03 \times 10^6 T/s

Explanation:

As we know that the power emitted by the source is given as

P = 170 W

now we know that

P = \frac{N}{t} (\frac{hc}{\lambda})

now we know that energy density is given as

u = \frac{B^2}{2\mu_0} + \frac{\epsilon_0 E^2}{2}

now we have

E = B c

u = \frac{B^2}{2\mu_0}

intensity is defined as

I = \frac{P}{A}

now we have

\frac{I}{c} = u = \frac{B^2}{2\mu_0}[/tex]

now we have

\frac{dB}{dt} = \omega B

\frac{dB}{dt} = \frac{2\pi c B}{\lambda}

\frac{dB}{dt} = \frac{2\pi c \sqrt{2\mu_0 I}}{\lambda\sqrt c}

here we have

I = \frac{P}{4\pi r^2}

I = \frac{170}{4\pi (410)^2}

I = 8.05 \times 10^{-5}

now we have

\frac{dB}{dt} = \frac{2\pi\sqrt{2\mu_0 c (8.05 \times 10^{-5})}}{(510 nm)}

\frac{dB}{dt} = 3.03 \times 10^6 T/s

4 0
3 years ago
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