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3 years ago

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An electron traveling horizontally to the right enters a region where a uniform electric field is directed downward. What is the

direction of the electric force exerted on the electron once it has entered the electric field?

1 answer:

astraxan [27]3 years ago

8 0

Answer:

Upward

Explanation:

For charged particles immersed in an electric field:

- if the particle is positively charged, the direction of the force is the same as the direction of the electric field

- if the particle is negatively charged, the direction of the force is opposite to the direction of the electric field

In this problem, we have an electron - so a negatively charged particle - so the direction of the force is opposite to that of the electric field.

Since the electric field is directed downward, therefore, the electric force on the electron will be upward.

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The air in a car tire la compressed when the car rolls over a rock. If the air

stealth61 [152]

Answer:

the signs of heat and work are; -Q and -W

Explanation:

The first law of thermodynamics is given by; ΔU = Q − W

where;

ΔU is the change in internal energy of a system,

Q is the net heat transfer (the sum of all heat transfer into and out of the system)

W is the net work done (the sum of all work done on or by the system).

Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).

Since work is done by the system, W remains negative.

Thus, the signs of heat and work are; -Q and - W

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3 years ago

A 5 kg fish swimming at 1 m/s swallows an absentminded 500 g fish swimming toward it at a velocity that brings both fish to a ha

AlexFokin [52]

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum is defined as the product between mass and velocity of each body. And its conservation as the equality between the initial and final momentum. Mathematically described as

$m_1u_1+m_2u_2 = (m_1+m_2)v_f$

Here

$m_1$ = Mass of big fish

$m_2$ = Mass of small fish

$v_1$ = Velocity of big fish

$v_2$ = Velocity of small fish

$v_F$ = Final Velocity

The big fish eats small fish and the final velocity is zero. Rearrange the equation for the initial velocity of small fish we have

$m_1u_1=-m_2u_2$

$u_2 = -\frac{m_1u_1}{m_2}$

Replacing we have,

$u_2 = -\frac{(5kg)(1m/s)}{0.5kg}$

$u_2 = -10m/s$

The negative sign indicates that the small fish is swimming in the direction opposite to that of the big fish.

Therefore the speed of the small fish is 10m/s

8 0

3 years ago

tickets at a museum cost 17 dollars each for a field trip, the museum offers a 4 doller discount on each ticket. How much will t

Marizza181 [45]

Answer:

$416 sounds like the best answer.

Explanation:

The tickets started off at $17, but a $4 discount for EACH ticket bought.

so 17-4 is 13

now tickets cost $13 each.

Multiply it by 32.

13x32=416

5 0

3 years ago

While doing a lab, a student found the density of a piece of pure aluminum to be 2.85 g/cm3. The accepted value for the density

bezimeni [28]

The equation for percent error is

% Error = $100*|Experimental-Theoretical|/Theoretical$

Our experimental is 2.85g/cm^3 and the accepted is 2.7g/cm^3

Thus our % Error = 5.555%

3 0

3 years ago

A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1

aivan3 [116]

Answer:

Part a)

$F_n = 306 N$

Part b)

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

$mg - F_n = ma_c$

now we will have

$mg - F_n = \frac{mv^2}{R}$

$F_n = mg - \frac{mv^2}{R}$

now we have

$F_n = 100(9.81) - \frac{100(9^2)}{12}$

$F_n = 981 - 675$

$F_n = 306 N$

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

$F_n + mg = ma_c$

$F_n = 0$ at minimum speed

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = \sqrt{15 \times 9.81}$

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

5 0

3 years ago