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Lerok [7]
2 years ago
14

A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106

meters/second, what is the force acting on the electrons? The value of q = -1.6 × 10-19 coulombs.
Physics
1 answer:
defon2 years ago
3 0

Answer:

4.7\cdot 10^{-14}N

Explanation:

For a charge moving perpendicularly to a magnetic field, the force experienced by the charge is given by:

F=qvB

where

q is the magnitude of the charge

v is the velocity

B is the magnetic field strength

In this problem,

q=1.6\cdot 10^{-19} C

v=6.5\cdot 10^6 m/s

B=4.5\cdot 10^{-2} T

So the force experienced by the electrons is

F=(1.6\cdot 10^{-19}C)(6.5\cdot 10^6 m/s)(4.5\cdot 10^{-2} T)=4.7\cdot 10^{-14}N

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The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a
Amanda [17]

Answer:

amount of energy  = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

so

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

payback period = \frac{excess initial cost}{\Delta M}

payback period = \frac{32 + 66}{520.34}

payback period = 0.188 year

8 0
2 years ago
An air-filled capacitor is formed from two long conducting cylindrical shells that are coaxial and have radii of 30 mm and 80 mm
Licemer1 [7]

Answer:

24

Explanation:

4 0
2 years ago
Jim stands beside a wide river and wonders how wide it is. he spots a large rock on the bank directly across from him. he then w
LuckyWell [14K]

To solve this problem, we must imagine that Jim’s initial position, the position of the rock, and Jim’s final position all connects to form a triangle. Now we can imagine that the triangle is a right triangle with the 90° angle on the initial position.

The angle of 30° is directly opposite to the length of his total stride while the width of the river is the side adjacent to the angle. Therefore can use the tan function to solve for the width of the river:

tan θ = opposite side / adjacent side

tan 30 = total stride distance / width of river

where total stride distance = 65 * 0.8 = 52 m

width of river = 52 m / tan 30

<span>width of river = 90.07 m</span>

7 0
2 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
1 year ago
What is meant in astronomy by the phrase "adaptive optics?
Sergeu [11.5K]
<h2>Answer: a.The mirrors and eyepiece of a large telescope are spring-loaded to allow them to return quickly to a known position. </h2>

Explanation:

Adaptive optics is a method used in several astronomical observatories to counteract in real time the effects of the Earth's atmosphere on the formation of astronomical images.

This is done through the insertion into the optical path of the telescope of sophisticated deformable mirrors supported by a set of computationally controlled actuators. Thus obtaining clear images despite the effects of atmospheric turbulence that cause the unwanted distortion.

It should be noted that with this technique it is also necessary to have a moderately bright reference star that is very close to the object to be observed and studied. However, it is not always possible to find such stars, so a powerful laser beam is used to point towards the Earth's upper atmosphere and create artificial stars.

7 0
2 years ago
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