1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
DaniilM [7]
2 years ago
6

When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is direc

ted:_____
a. into the page.
b. toward the left
c. toward the right
d. toward the bottom of the page.
e. toward the top of the page.
f. out of the page.
Physics
1 answer:
Serga [27]2 years ago
8 0

Answer: F

Out of the page.

Explanation:

For an electron with a charge of -e, the magnitude of the force on it is F = BeV

Where

F = force on the electron

e = charge ( electrons )

V = velocity

B = magnetic field

F is the force acting on all the electrons in a wire which gives rise to the F = BIL

Where

I = current

L = length of the wire

The force F is always at the right angle to the particle's velocity and its direction can be found using the left hand rule.

When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.

You might be interested in
If a ball that is 10 meters above the ground is thrown horizontally at 5.51 meters per second. a. how long will it take for the
GalinKa [24]

Answer:

a. t = 1.43 s

b. d = 7.88 m

Explanation:

a. The time of flight can be found using the following equation:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height = -10 m

y_{0}: is the initial height = 0

v_{0_{y}}: is the initial speed in the vertical direction = 0

g: is the acceleration due to gravity = 9.81 m/s²

By solving the above equation for "t" we have:

t = \sqrt{\frac{2y_{f}}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s

Hence, the ball will hit the ground in 1.43 s.

b. The distance in the horizontal direction can be found as follows:

x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2}

Where:

x₀: is the initial position in the horizontal direction = 0

a: is the acceleration in the horizontal direction = 0 (it is moving at constant speed)

x_{f} = 5.51 m/s*1.43 s = 7.88 m

Therefore, the ball will travel 7.88 m before it hits the ground.

I hope it helps you!

4 0
2 years ago
This back-and-forth movement of electrons is called . In contrast, the movement of electrons in one direction in a battery circu
stich3 [128]
The back-and-forth movement of electrons is called alternating current. Electrons go back and forth, the direction of their path alternates from one direction to another.

the movement of electrons in one direction is called direct current. The electrons move in a direct, single path without changing directions.
5 0
3 years ago
What is the height of the image? Round the answer
Maru [420]
It’s a concave mirror
8 0
2 years ago
A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away
Artist 52 [7]

Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

d)  Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it.  Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball.   Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.  

   Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

                               y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

                              0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

                               V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

                               x = x(0) + V_i,x*t_c

                               D = 0 + V_i,x*t_c

                               V_o,x + v_r = D/t_c

                                V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

                           V_o = sqrt ( V_o,x^2 + V_o,y^2)

                          V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

       

- The angle with which it should be thrown can be evaluated by trigonometric relation:

                            tan(Q) = ( V_o,y / V_o,x )

                            tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

                                   Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

                           

                               

6 0
2 years ago
A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
stealth61 [152]

Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

5 0
3 years ago
Other questions:
  • For a solution to be a physical solution, it must satisfy several criteria. First, it must be continuous everywhere. Second, it
    9·1 answer
  • What conditions must be met if a man-made satellite is to attain a circular orbit around the earth?(choose as many as apply)
    15·1 answer
  • Explain the human impact on local lakes and ponds, such as Rough River, Nolin, or Caneyville Watershed. Answer in 6-8 sentences
    9·2 answers
  • Calculate the potential difference across a 25- resistor if a 0.3-A current is flowing through it.
    6·2 answers
  • How do you derive the drift velocity equation (I = nAve)?
    11·1 answer
  • Place these bodies of our solar system in the proper order of formation
    5·2 answers
  • A hoop of mass 2 kg, radius 0.5 m is rotating about its center with an angular speed of 3 rad's. A force of 10N is applied tange
    8·1 answer
  • g Determine the magnetic field midway between two long straight wires 2.0 cm apart in terms of the current I in one when the oth
    10·1 answer
  • You are kicking your soccer ball back and forth on a soccer field. You kick the ball forward for 10 m. You turn completely aroun
    14·1 answer
  • Difference between 3.15 m and 2.0 m with the correct number of significant figures?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!