All categories

2 years ago

1000 millicoulombs of charge passes through a point. The amount of current passing through the point is

Physics

1 answer:

Rzqust [24]2 years ago

3 0

The amount of current passing through the point is 1 A

The amount of current passing through the point can be calculated using the formula below.

⇒ Formula:

Q = i/t......................... Equation 1

⇒ Where:

Q = Charge
i = current
t = time.

⇒ Make "i" the subject of the equation.

i = Qt....................... Equation 2

From the question,

⇒ Given:

Q = 1000 millicoulombs = 1 coulombs
t = 1 seconds. (Assuming the time is 1 seconds)

⇒ Substitute these values into equation 2

i = 1/1
i = 1 A.

Hence, The amount of current passing through the point is 1 A.

Learn more about charges here: brainly.com/question/4158552

You might be interested in

Need amswer fast plz

FrozenT [24]

Answer:

I need this for may schooling

6 0

2 years ago

I NEED AN ANSWER!

Lera25 [3.4K]

Answer:

Before sled starts to move it has a potential energy due to the elevation...and then that potential energy converted to kinetic energy due to presence of a velocity...the sled will continue to move if their is no resesive force...but however friction force is presence that cause the sled to stop....

5 0

2 years ago

A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.2 meters/second. At what horizontal distance fr

NARA [144]

To find the horizontal distance multiple the horizontal velocity by the time. Since there is no given time it must be calculated using kinematic equation.

Y=Yo+Voyt+1/2at^2
0=.55+0+1/2(-9.8)t^2
-.55=-4.9t^2
sqrt(.55/4.9)=t
t=0.335 seconds

Horizontal distance
=0.335s*1.2m/s
=0.402 meters

8 0

3 years ago

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

2 years ago

A track athlete jumps over a hurdle. What is the equal and opposite force for the force of the athlete pushing on the ground as

Jet001 [13]

Answer: Normal Force of the ground pushing the athlete up.

Explanation: I did this on Khan Academy and got the answer right.

6 0

3 years ago