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Advocard [28]
3 years ago
13

An electric field can be created by a single charge or a distribution of charges. The electric field a distance from a point cha

rge has magnitude E = k|q'|/r^2.The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point charges. Note that if a charge is placed in an electric field created by q', q will not significantly affect the electric field if it is small compared to q'.Imagine an isolated positive point charge with a charge Q (many times larger than the charge on a single electron).There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend?Check all that apply.A the distance between the positive charge and the electronB the charge on the electronC the mass of the electronD the charge of the positive chargeE the mass of the positive chargeF the radius of the positive chargeG the radius of the electron
Physics
1 answer:
joja [24]3 years ago
8 0

Answer:

A the distance between the positive charge and the electron

B the charge on the electron

D the charge of the positive charge

Explanation:

The electric field produced by the positive charge Q at the location of the electron is given by

E=k\frac{Q}{r^2}

where

k is the Coulomb constant

Q is the charge

r is the distance between the charge Q and the electron

The force exerted on a charged particle by an electric field is given by

F=qE

where q is the magnitude of the charged particle. So, the force exerted on the electron in this problem is

F=eE = k\frac{eQ}{r^2}

where e is the charge of the electron. As we see from the equation, the force depends only the following quantities:

A the distance between the positive charge and the electron (r)

B the charge on the electron (e)

D the charge of the positive charge (Q)

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A hockey puck slides off the edge of a platform with an initial velocity of 20 m/s horizontally. The height of the platform abov
Rina8888 [55]

Answer:

20.96 m/s

Explanation:

Using the equations of motion

y = uᵧt + gt²/2

Since the puck slides off horizontally,

uᵧ = vertical component of the initial velocity of the puck = 0 m/s

y = vertical height of the platform = 2 m

g = 9.8 m/s²

t = time of flight of the puck = ?

2 = (0)(t) + 9.8 t²/2

4.9t² = 2

t = 0.639 s

For the horizontal component of the motion

x = uₓt + gt²/2

x = horizontal distance covered by the puck

uₓ = horizontal component of the initial velocity = 20 m/s

g = 0 m/s² as there's no acceleration component in the x-direction

t = 0.639 s

x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m

For the final velocity, we'll calculate the horizontal and vertical components

vₓ² = uₓ² + 2gx

g = 0 m/s²

vₓ = uₓ = 20 m/s

Vertical component

vᵧ² = uᵧ² + 2gy

vᵧ² = 0 + 2×9.8×2

vᵧ = 6.26 m/s

vₓ = 20 m/s, vᵧ = 6.26 m/s

Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s

4 0
3 years ago
Read 2 more answers
How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?
Anestetic [448]

Answer:

21 m

Explanation:

The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

d=vt

where

d is the distance covered

v is the speed

t is the time

The frog in this problem has a speed of

v = 2.1 m/s

and therefore, after t = 10 s, the distance it covered is

d=(2.1)(10)=21 m

3 0
4 years ago
At a particular instant the magnitude of the momentum of a planet is 2.60 × 10^29 kg·m/s, and the force exerted on it by the sta
aleksley [76]

Answer:

F=(-4.8*10^22,0,0) N

Explanation:

<u>Given  :</u>

We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°

Solution :

We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.  

The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next  

F=|F|cosФp            (1)

Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet  

F=|F|cosФp

 =-4.8*10^22 N*p

<em>As this force is in one direction, we could get its vector as next  </em>

F=(-4.8*10^22,0,0) N

F=(0,-4.8*10^22,0) N

F=(0,0-4.8*10^22) N

The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.

8 0
3 years ago
which of the following is accurate in describing the placement and classification of iodine? a. iodine is located in period 5, g
Elden [556K]
A. iodine is located in period 5, group 17 and is classified as a nonmetal

Group is the column, period is the raw.

8 0
3 years ago
To solve a problem using the equation for keplerâs third law, enrico must convert the average distance of mars from the sun from
Natali5045456 [20]
1 astronomical unit = 149597870700m
Enrico should divide distance in meters with this number.
3 0
3 years ago
Read 2 more answers
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