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Yuki888 [10]
3 years ago
14

7. A neutral aluminum rod is at rest on a foam insulating base. A negatively charged balloon is brought near one end of the rod

but not in direct contact with it. In what way, if any, will the charges in the rod be affected?
Physics
1 answer:
WITCHER [35]3 years ago
6 0

Answer:

If a negatively charged balloon is brought near one end of the rod but not in direct contact, then <u>the negative charges on the balloon repel the same amount of negative charges on the end of the rod that is close to the balloon</u>, and the positive charges stay at the balloon-side of the rod. The total charge of the rod is still zero, but the distribution of the charges are now non-uniform.

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When is the velocity of a mass on a spring at its maximum value?
ehidna [41]

Answer:

A.  when the mass has a displacement of zero

Explanation:

The velocity of a mass on a spring can be calculated by using the law of conservation of energy. In fact, the total energy of the mass-spring system is equal to the sum of the elastic potential energy (U) of the spring and the kinetic energy (K) of the mass:

E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2

where

k is the spring constant

x is the displacement of the mass with respect to the equilibrium position of the spring

m is the mass

v is the velocity of the mass

Since the total energy E must remain constant, we can notice the following:

- When the displacement is zero (x=0), the velocity must be maximum, because U=0 so K is maximum

- When the displacement is maximum, the velocity must be minimum (zero), because U is maximum and K=0

Based on these observations, we can conclude that the velocity of the mass is at its maximum value when the displacement is zero, so the correct option is A.


8 0
3 years ago
Read 2 more answers
g Estimate the number of photons emitted by the Sun in a second. The power output from the Sun is 4 × 1026 W and assume that th
vagabundo [1.1K]

Answer:

The value is N  =  1.107 *10^{45 }  \ photons    

Explanation:

From the question we are told that

   The  power output from the sun is  P_o =  4 * 10^{26} \  W

   The average wavelength of each photon is  \lambda  = 550 \  nm  =  550 *10^{-9} \  m

Generally the energy of each photon emitted is mathematically represented as

        E_c =  \frac{h * c  }{ \lambda }

Here  h is the Plank's constant with value  h  =  6.62607015 * 10^{-34} J \cdot s

          c is the speed of light with value  c =  3.0 *10^{8} \  m/s

So

       E_c =  \frac{6.62607015 * 10^{-34}  * 3.0 *10^{8}  }{ 550 *10^{-9} }          

=>   E_c =  3.614 *10^{-19} \  J          

Generally the  number of photons emitted by the Sun in a second is mathematically represented as

         N  =  \frac{P }{E_c}

=>      N  =  \frac{4 * 10^{26} }{3.614 *10^{-19}}

=>      N  =  1.107 *10^{45 }  \ photons    

5 0
2 years ago
What is the force of gravity (from the Earth) on the 700kg satellite if it’s 10km above the Earths surface?
joja [24]

Answer:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

    = {\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2} = 3984378 m / s^{2}

Explanation:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

    = {\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2} = 3984378 m / s^{2}

3 0
3 years ago
What happens to light waves at the interface between different media?
nevsk [136]

Answer:

Refractive motion is the impact of a light wave that travels from medium to medium in an angle away from normal, where the direction of light varies. Light is refracted when it crosses the air-to-glass interface and moves slower.

Explanation:

Refractive motion is the impact of a light wave that travels from medium to medium in an angle away from normal, where the direction of light varies. Light is refracted when it crosses the air-to-glass interface and moves slower.

Hope this helps.

6 0
3 years ago
An electron and a proton are fixed at a separation distance of 973 nm. Find the magnitude and the direction of the electric fiel
Vilka [71]

Answer:

The magnitude is: |E|=6084.1\: N/C

The direction of E is in the negative x-direction.

Explanation:

The electric field equation is:

E=k\frac{Q}{r^{2}}

Where:

  • Q is the charge (we can choose the electron or the proton)
  • r is the distance (in our case is at the midpoint 973/2 nm)
  • k is the Coulomb constant (9*10^{9}\: Nm^{2}C^{-2})

Using the electron charge (e = -1.6*10^{-19}\: C)

E=-9*10^{9}\frac{1.6*10^{-19}}{(486.5*10^{-9})^{2}}

The magnitude is:

|E|=6084.1\: N/C

The direction of E is in the negative x-direction.

I hope it helps you!

6 0
2 years ago
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