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Wewaii [24]
1 year ago
6

A beam of electrons is sent horizontally down the axis of a tube to strike a fluorescent screen at the end of the tube. On the w

ay, the electrons encounter a magnetic field directed vertically downward. The spot on the screen will therefore be deflected: 1. not at all 2. upward 3. to the right as seen from the electron source 4. to the left as seen from the electron source 5. downward
Physics
1 answer:
slamgirl [31]1 year ago
4 0

Answer:

The answer is 3.

Explanation:

The answer to this question can be found by applying the right hand rule for which the pointer finger is in the direction of the electron movement, the thumb is pointing in the direction of the magnetic field, so the effect that this will have on the electrons is the direction that the middle finger points in which is right in this example.

So as a result of the magnetic field directed vertically downwards which is at a right angle with the electron beams, the electrons will move to the right and the spot will be deflected to the right of the screen when looking from the electron source.

I hope this answer helps.

You might be interested in
When a gas is rapidly compressed (say, by pushing down a piston) its temperature increases. When a gas expands against a piston,
shusha [124]

Answer:

Explained in explanation

Explanation:

The first law of thermodynamics states that the change in internal energy of a system(ΔU) is equal to the sum of the net heat transfer into the system(Q) and the net work done on the system(W). In equation, this law is;

ΔU = Q + W

Now, when there's gas inside a container with a movable piston that's tightly fitting, we will assume that the piston can move up and down thereby compressing the gas or allowing the gas to expand against it.

Now these gas molecules inside the container possess kinetic energy. Thus, the internal energy(U) of the system is simply the sum of all the kinetic energies of the individual gas molecules present in the container.

Therefore, if the temperature(T) of the gas increases, then the speed and internal energy(U) of the gas molecules will also increase. In the same way, if the temperature of the gas decreases, the speed and internal energy of the gas molecules would also decrease.

Now, back to the question, when the piston is pushed down, it does work on the gas and the gas does negative work on the piston. Thus, the gas will be get compressed to a smaller space, and thereby making the gas molecules to hit the piston at a faster rate. Thus, there is a decrease in speed and as we saw earlier that when there is a decrease in speed, it means temperature has decreased.

Whereas, when the piston is moved up, the gas does positive work on the piston and the speed of the gas molecules will increase. Like I said earlier that increase in speed means increase in temperature.

4 0
1 year ago
You buy a AA battery in the store, and it is marked 1.5 V. If this marking is strictly accurate, while this battery is fresh its
vitfil [10]

Answer:

Option A is correct.

when it is used in a circuit. its terminal voltage will be less than 1.5 V.

Explanation:

The terminal voltage of the battery when it is in use in circuits drops lower than the 1.5 V rating given to it due to internal resistance.

All batteries give internal resistances when used in circuits. The internal resistance (though very small) is usually modelled as connected in series with the battery. It is due to some form of interference from the chemical makeup of the battery.

Normally, while the battery is fresh, the voltage (V) obtained at its terminals when connected in series with a resistor of resistance R is V = IR; where I is the current flowing in this circuit.

But once the interenal resistance (r) of the battery comes into play,

V = I₁ (r + R)

The current in the circuit evidently drops (that is I₁ < I) and V = (I₁r + I₁R)

The voltage across the terminals of the battery is no longer V but is now (V) × [R/(R+r)] which is less than the initial V and it reduces as the internal resistance, r, increases.

Hope this Helps!!!

3 0
2 years ago
The wind blows a jay bird south with a force of 300 Newtons. The
adoni [48]

Answer:

F = 316.22 N

Explanation:

Given that,

The wind blows a jay bird south with a force of 300 Newtons.

The  jay bird flies north, against the wind, with a force of 100  newtons.

Both the forces are acting perpendicular to each other. The net force is given by the resultant of forces as follows :

F=\sqrt{300^2+100^2} \\\\F=316.22\ N

Hence, the net force on the jay bird is 316.22 N.

6 0
2 years ago
A pesticide was applied to a population of roaches, and it was determined that the LD50 was 55mgkg. If the average mass of a roa
zhannawk [14.2K]

Answer:

55mg/kg * 0.02kg which gives 1.1 mg pesticides

Explanation:

(Options are missing)

If the average mass or weight of a roach is 0.02kg

And 55mg is needed per 1 kg roaches

The amount or size of pesticide required for 0.02kg roach is 55mg/kg * 0.02kg

Size = 55mg/kg * 0.02 kg

Size = 1.1mgkg/kg

Size = 1.1 mg

So, the size of pesticide required is 1.1mg

0 0
1 year ago
A boy throws a water ballon such that it hits his sister standing 10m away. The boy threw the water balloon at an angle of 35 de
Tresset [83]

Answer: 7.734 m/s

Explanation:

We have the following data:

\theta=35\° The angle at which the water ballon was thrown

x=10 m  The horizontal distance of the water ballon

g=-9.8 m/s^{2} The acceleration due gravity

We need to find the initial velocity V_{o} at which the water ballon was thrown, and we can find it by the following equation:

x=V_{o}cos \theta T (1)

Where T=2t is the total time the water ballon is on air

On the other hand, when we talk about parabolic motion (as in this situation) the water ballon reaches its maximum height just in the middle of this parabola, when V=0 and the time t is half the time T it takes the complete parabolic path.

So, if we use the following equation, we will find t:

V=V_{o}+gt=0 (2)

Isolating t:

t=\frac{-V_{o}}{g} (3)

Remembering T=2t:

T=2\frac{-V_{o}}{g} (4)

Substituting (4) in (1):

x=V_{o}cos \theta (2\frac{-V_{o}}{g}) (5)

Isolating V_{o}:

V_{o}=\sqrt{\frac{x g}{-2 cos \theta}} (6)

V_{o}=\sqrt{\frac{(10 m)(9.8 m/s^{2})}{-2 cos(35\°)}} (7)

Finally:

V_{o}=7.734 m/s

4 0
1 year ago
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