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Vsevolod [243]
3 years ago
8

How would the path of the ball differ on Earth than on the moon?Which statement best compares the accelerations of two objects i

n free fall?
Physics
1 answer:
TEA [102]3 years ago
8 0

Range on earth and height reached is less than on the moon.

accn on moon less than that on earth

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An electron that has an instantaneous velocity of ???? = 2.0 × 106 m ???? ???? + 3.0 × 106 m ???? ???? is moving through the uni
Setler79 [48]

Explanation:

It is given that,

Velocity of the electron, v=(2\times 10^6i+3\times 10^6j)\ m/s

Magnetic field, B=(0.030i-0.15j)\ T

Charge of electron, q_e=-1.6\times 10^{-19}\ C

(a) Let F_e is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

F_e=q_e(v\times B)

F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)]

F_e=-1.6\times 10^{-19}\times (-390000)(k)

F_e=6.24\times 10^{-14}k\ N

(b) The charge of electron, q_p=1.6\times 10^{-19}\ C

The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.

8 0
3 years ago
A student connects a small solar panel to a 40 a resistor to make a simple circuit. The solar panel produces a voltage of 2 0.
Nastasia [14]
<h3>Solution for the above question : -</h3>

Ohm's law states that :

  • v = ir

the terms used are :

  • r = resistance
  • v = potential \:  \: difference
  • i  = current

let's solve for electric current :

  • 2 = i \times 40

  • i =  \dfrac{2}{40}

  • i = 0.05 \: A

  • i = 50 \: mA

\mathfrak{good\:  \: luck \:  \: for \:  \: your \:  \: assignment}

8 0
2 years ago
The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 sec
marin [14]

Answer:

The additional trials needed is 48 trials

Explanation:

Given;

initial number of trials, n = 16 trials

the standard deviation, σ = 0.24 s

initial standard error, ε = 0.06 s

The standard error is given by;

\epsilon = \frac{\sigma}{\sqrt{n} }

To reduce the standard error to 0.03 s, let the additional number of trials = x

0.03= \frac{0.24}{\sqrt{n+x} } \\\\0.03= \frac{0.24}{\sqrt{16+x} }\\\\0.03\sqrt{16+x} = 0.24\\\\\sqrt{16+x} = \frac{0.24}{0.03} \\\\\sqrt{16+x} = 8\\\\16+x = 8^2\\\\16+x = 64\\\\x = 64 -16\\\\x = 48 \ trials

Therefore, the additional trials needed is 48 trials.

6 0
3 years ago
If two particles have equal kinetic energies, are their momenta necessarily equal? explain.
Mandarinka [93]

Answer:

No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

K.E=\frac{1}{2}mv^{2}

Similarly the momentum is given by m\times v

For 2 particles with masses m_{1},m_{2}and moving with velocities v_{1},v_{2} respectively the respective kinetic energies is given by

K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}

K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}

Similarly For 2 particles with masses m_{1},m_{2}and moving with velocities v_{1},v_{2} respectively the respective momenta are given by

p_{1}=m_{1}\times v_{1}

p_{2}=m_{2}\times v_{2}

Now since it is given that the two kinetic energies are equal thus we have

\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

5 0
3 years ago
The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the wave shown?
belka [17]

Answer:

B) 0.3Hz

Explanation:

I just took the test i hope i helped and i hope you pass the test

7 0
3 years ago
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