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olya-2409 [2.1K]
3 years ago
15

Which term best describes how fast and in what direction an object is moving at a particular moment? Question 2 options: acceler

ation average velocity position instantaneous velocity
Physics
1 answer:
kompoz [17]3 years ago
7 0

Answer:

Instantaneous velocity

Explanation:

Acceleration = rate of change of velocity

Average velocity = total displacement divided by total time taken

Position = It shows the location of the object

Instantaneous velocity = It is the velocity of an object at a particular instant.

The term that describes how fast and in what direction an object is moving at a particular moment is "instantaneous velocity". It is basically equal to the derivative of position wrt time.

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A current of 3.60A flows for 15.3s through a conductor. Calculate the number of electrons that pass through a point in the condu
Artyom0805 [142]

3.60 A = 3.60 coulombs of charge per second

(3.60 Coul/sec) x (15.3 sec) = 55.08 coulombs of charge

1 coulomb of charge is carried by 6.25 x 10^18 electrons

Number of electrons =

               (55.08 Coul) x (6.25 x 10^18 e/coul) = <em>3.4425 x 10^20 electrons</em>


7 0
3 years ago
a ball of mass 0.80 kg moving at a speed of 2.5 m/s along a straight line collided with a mass 2.5 kg which was initially statio
Likurg_2 [28]

The speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

The given parameters;

  • mass of the ball, m₁ = 0.8 kg
  • speed of the ball, u₁ = 2.5 m/s
  • mass of the object at rest, m₂ = 2.5 kg
  • final velocity of the object at rest, v₂ = 1 m/s

Let the final velocity of the 0.8 kg ball immediately after collision = v₁

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁  +  m₂v₂

(0.8 x 2.5) + (2.5 x 0) = (0.8)v₁  +  2.5(1)

2 = 2.5 + (0.8)v₁

-0.5 = (0.8)v₁

v_1 = \frac{-0.5}{0.8} \\\\v_1 = -0.625 \ m/s

Thus, the speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

Learn more here: brainly.com/question/7694106

5 0
2 years ago
In a bicycle race on a straight road, the leader is 72.0 m from the finish line and continues to travel to the finish line with
dem82 [27]

Try this solution, answers are marked with red colour.

8 0
3 years ago
Newton's First and Second Laws of Motion:Question 3Kepler-10b, the first confirmed terrestrial extrasolar planet, is about 564 l
OleMash [197]

Answer:

F = 85696.5 N = 85.69 KN

Explanation:

In this scenario, we apply Newton's Second Law:

Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\

where,

F = Upthrust = ?

m = mass of space craft = 5000 kg

g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)

g = 15.0093 m/s²

a = acceleration required = 2.13 m/s²

Therefore,

F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\

<u>F = 85696.5 N = 85.69 KN</u>

8 0
3 years ago
(a) Find the frequency of revolution of an electron with an energy of 114 eV in a uniform magnetic field of magnitude 46.7 µT. (
stira [4]

Answer:

(a)  1.3 x 10^6 Hz

(b) 76.73 cm

Explanation:

(a)

the formula for the frequency is given by

f = B q / 2 π m

where, B be the strength of magnetic field, q be the charge on one electron, m is the mass of one electron.

B = 46.7 micro tesla = 46.7 x 10^-6 T

q = 1.6 x 10^-19 C

m = 9.1 x 10^-31 kg

f = (46.7 x 10^-6 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31) = 1.3 x 10^6 Hz

(b) K = 114 eV = 114 x 1.6 x 10^-19 J = 182.4 x 10^-19 J

K = 1/2 mv^2

182.4 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v = 6.3 x 10^6 m/s

r = m v / B q

Where, r be the radius of circular path

r = (9.1 x 10^-31 x 6.3 x 10^6) / (46.7 x 10^-6 x 1.6 x 10^-19)

r = 0.7673 m = 76.73 cm  

5 0
3 years ago
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