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3 years ago

What is the acceleration of a proton moving with a speed of 9.5 m/s at right angles to a magnetic field of 1.8 T

Physics

1 answer:

Elenna [48]3 years ago

4 0

Explanation:

F = qvBsin(angle)

F = 1.6 x 10^-19 x 9.5 x 1.8 x Sin (90)

F = 273.6 x 10^-20N

a = F/M

M of a proton = 1.67 x 10^-27kg

a = 273.6 x 10^-20/1.67 x 10^-27

a = 1638323353.29m/s

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Precautions taken when doing experiment on simple pendulum

olchik [2.2K]

Answer:

1.)The bob of pendulum should be displaced with a small angle.

2) The amplitude of the oscillation of a simple pendulum should be small.

3) Fans should be switched off to reduce the air resistance.

4)The simple pendulum should be oscillate in a vertical plane

Hope this helped you

7 0

4 years ago

Two hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area 1.20 × 10−3 m2

Novosadov [1.4K]

Answer:

The charge flows through a point in the circuit during the change is 0.044 C.

Explanation:

Given that,

Number of turns in the copper wire, N = 200

Area of cross section, $A=1.2\times 10^{-3}\ m^2$

Resistance of the circuit, R = 118 ohms

If an externally applied uniform longitudinal magnetic filed in the core changes from 1.65 T in one direction to 1.65 T in the opposite direction.

We need to find the charge flows through a point in the circuit during the change. Due to change in magnetic field an emf is induced in it. It is given by :

$\epsilon=\dfrac{d\phi}{dt}$

Using Ohm's law :

$\epsilon=IR$

$IR=-\dfrac{d\phi}{dt}\\\\I=-\dfrac{1}{R}\dfrac{d\phi}{dt}$

Electric current is equal to the rate of change of electric charge. So,

$dq=\dfrac{NA(B(0)-B(t))}{R}\\\\dq=\dfrac{200\times 1.2\times 10^{-3}(1.65+1.65)}{18}\\\\dq=0.044\ C$

So, the charge flows through a point in the circuit during the change is 0.044 C.

4 0

3 years ago

What provided the force that made the cart speed up

fgiga [73]

Answer:

Potential energy turn to kinetic energy

Explanation:

5 0

4 years ago

Read 2 more answers

A weightlifter must exert 25 Newtons of force to

kakasveta [241]

$Work=Force \: × \: displacement \\ => W=Fs \\ \\ Given, \\ Force=25 \: N \\ Displacement=2m \\ \\ so \: work = 25 \: N \: \times 2m \\ = > work = 25kgm {s}^{ - 2} \times 2m \\ = > work = 50kg {m}^{2} {s}^{ - 2} \\ = > work = 50J$

This is the answer.

Hope it helps!!

6 0

3 years ago

The electric potential at a certain distance from a point charge can be represented by V. What is the value of the electric pote

Maksim231197 [3]

Answer:

<em>b. At twice the distance, the electric potential is V/2</em>

Explanation:

Electric potential

Is the amount of work needed to move a unit charge from a reference point (usually a point at infinity) without producing acceleration.

The electric potential due to a point charge q at a distance r is given by

$\displaystyle V=K\frac{q}{r}$

Where K is the Coulomb's constant. If we know the electric potential at a certain distance is V, if the distance is changed to 2r, then the new potential is

$\displaystyle V'=K\frac{q}{2r}=\frac{1}{2}K\frac{q}{r}=\frac{1}{2}V$

It means that the electric potential is half the previous value. Correct option: b.

8 0

3 years ago