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
Troyanec [42]
3 years ago
9

A charge q moves from point A to point B. The potential energy of the charge at point A is 5.3 _ 10-10 joules and at point is B

is 3.1 _ 10-10 joules. If the kinetic energy of the charge at point A is zero, what is the kinetic energy at point B?
Select one of the options below as your answer:

A.
8.4 _ 10-10 joules
B.
5.3 _ 10-10 joules
C.
3.1 _ 10-10 joules
D.
2.2 _ 10-10 joules
E.
1.5 _ 10-10 joules
Physics
2 answers:
Sphinxa [80]3 years ago
8 0

D is the answer :D hope it helps

stiv31 [10]3 years ago
7 0
The change in electric potential energy will be converted to kinetic energy; thus:
K.E = 5.3 x 10⁻¹⁰ - 3.1 x 10⁻¹⁰
K.E = 2.2 x 10⁻¹⁰ Joules
Option D is correct.
You might be interested in
A device known as an optical resonator is used in lasers. An optical resonator consists of an arrangement of mirrors that reflec
polet [3.4K]

Answer:

A. The resonator behaves as a wave guide (a hollow pipe used as a transmission line). The characteristics of the pipe depend on the type of the wave to be transmitted.

4 0
2 years ago
A truck with 0.410 m radius tires travels at 25.0 m/s. What is the angular velocity of the rotating tires in radians per second?
Eduardwww [97]

relation between linear velocity and angular velocity is given as

v = R\omega

here

v = linear speed

R = radius

\omega = angular speed

now plug in all data in the equation

25.0 = 0.410 \omega

\omega  = \frac{25}{0.410}

\omega = 60.9 rad/s

so rotating speed is 60.9 rad/s

6 0
3 years ago
A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel
Dafna1 [17]

Answer:

1.69\cdot 10^{10}J

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

E=-G\frac{mM}{2r}

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

M=5.98\cdot 10^{24}kg is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m

So the initial total energy is

E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J

When the satellite hits the ground, it is now on Earth's surface, so

r=R=6370 km=6.37\cdot 10^6 m

so its gravitational potential energy is

U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J

And since it hits the ground with speed

v=1.90 km/s = 1900 m/s

it also has kinetic energy:

K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J

So the total energy when the satellite hits the ground is

E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J

8 0
3 years ago
Salmon often jump waterfalls to reach their
PilotLPTM [1.2K]

The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

  • height of the waterfall, h = 0.432 m
  • distance of the Salmon from the waterfall, s = 3.17 m
  • angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s

The minimum velocity of the Salmon jumping at the given angle is calculated as;

X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

Learn more here: brainly.com/question/20064545

8 0
2 years ago
Julia can swim at 3.5 km/h in still water. She attempts to head straight north
raketka [301]

1) 3.7 km/hr  N19°W

2) she must aim at N20°E

work sheet  #6

relative velocity

3 0
3 years ago
Other questions:
  • An open vertical tube has water in it. a tuning fork vibrates over its mouth. as the water level is lowered in the tube, a reson
    14·1 answer
  • Upon its return voyage from a space mission, the spacecraft has a velocity of 25000 km/h at point A, which is 7130 km from the c
    10·1 answer
  • A 8.0 n force acts on a 0.70-kg object for 0.50 seconds. by how much does the object's momentum change (in kg-m/s)? (never inclu
    6·1 answer
  • IN a physics lab, a student discovers that the magnitude of the magnetic field at a certain distance from a long wire is 4.0μT.
    10·1 answer
  • Based on the information in the table, which two elements are most likely in the same group, and why? bismuth and thallium, beca
    6·2 answers
  • What does balanced and unbalanced forces have to do with newtons laws of motion
    10·1 answer
  • A camera gives a proper exposure when set to a shutter speed of 1/250 s at f-number F8.0. The photographer wants to change the s
    8·1 answer
  • A student wants to determine the speed of sound at an elevation of one mile. To do this the student performs an experiment to de
    8·1 answer
  • Describe Newton's law of gravitation and its application?​
    12·1 answer
  • Arrange the balls in order from greatest amount of gravitational potential energy to least.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!