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2 years ago

What is the difference between a circuit and a current?

Physics

2 answers:

Mrrafil [7]2 years ago

6 0

Answer:

Electric current is the charge that flows through the circuit. Electric circuit is the path of conduction wires through which current flows.

Explanation:

hope this helps. have a good day!

umka2103 [35]2 years ago

6 0

a circuit is a complete loop that carries a current in the form of electrons from negative to positive.

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lonnie pitches a baseball of mass 0.02kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to L

vodomira [7]

Answer:

I=2 kg.m/s

Explanation:

The impulse is defined as the change of momentum:

$I=p_f-p_o\\I=m*v_f-m*v_o\\I=0.02kg*[(-60m/s)-40m/s]\\I=2kg.m/s$

We took the final velocity as negative since it is going on the opposite direction of the intial motion of the ball.

8 0

3 years ago

Two stars that are in the same constellation:

sammy [17]

Answer:

A binary star is a star system consisting of two stars orbiting around their common barycenter.

<h3>❣️(◍Jess bregoli◍)❣️</h3>

#keep learning

7 0

2 years ago

(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th

Ghella [55]

Answer:

Assuming that the vertical speed of the ball is 14 m/s we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

Explanation:

a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:

$V_{f}^{2} = V_{0}^{2} - 2gh$

Where:

$V_{f}$ : is the final speed = 14 m/s

$V_{0}$ : is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

$V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s$

b) The maximum height is:

$V_{f}^{2} = V_{0}^{2} - 2gh$

$h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m$

c) The time can be found using the following equation:

$V_{f} = V_{0} - gt$

$t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s$

d) The flight time is given by:

$t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s$

I hope it helps you!

3 0

3 years ago

What is yo main vein call????

mariarad [96]

Artery im pretty sure

7 0

3 years ago

Read 2 more answers

n the railroad accident, a boxcar weighting 200 kN and traveling at 3 m/s on horizontal track slams into a stationary caboose we

USPshnik [31]

Answer:

ΔK = -6 10⁴ J

Explanation:

This is a crash problem, let's start by defining a system formed by the two trucks, so that the forces during the crash have been internal and the moment is preserved

initial instant. Before the crash

p₀ = m v₁ + M 0

final instant. Right after the crash

p_f = (m + M) v

p₀ = p_f

mv₁ = (m + M) v

v = $\frac{m}{m+M} \ v_1$

we substitute

v = $\frac{20}{20+40}$ 3

v = 1.0 m / s

having the initial and final velocities, let's find the kinetic energy

K₀ = ½ m v₁² + 0

K₀ = ½ 20 10³ 3²

K₀ = 9 10⁴ J

K_f = ½ (m + M) v²

K_f = ½ (20 +40) 10³ 1²

K_f = 3 10⁴ J

the change in energy is

ΔK = K_f - K₀

ΔK = (3 - 9) 10⁴

ΔK = -6 10⁴ J

The negative sign indicates that the energy is ranked in another type of energy

7 0

3 years ago