All categories

3 years ago

What is the average electric current in a wire, when a charge of 150 C passes in 30 s?

1 answer:

7 0

We know, I = Q / t
Here, Q = 150 C
t = 30 s

Substitute their values,
I = 150 / 30
I = 5 A

In short, Your Answer would be 5 Ampere

Hope this helps!

You might be interested in

A 98-kg fullback, running at 5.0 m/s, attempts to dive directly across the goal line for a touchdown. Just as he reaches the lin

Troyanec [42]

Answer:

(a) Explained below

(b) $v_f=0.35\ m/s$

Explanation:

Law Of Conservation Of Linear Momentum

The total linear momentum of a system of particles or objects is conserved unless an external force is acting on the system. The formula for the momentum of a body with mass m and velocity v is P=mv. If there is a system of bodies, then the total linear momentum is the sum of the individual momentums

$P=m_1v_1+m_2v_2+...+m_nv_n$

When objects collide and join together, the only final mass is the sum of all masses, all traveling at the same speed.

Our $m_1=98\ kg$ fullback runs at $v_1=5\ m/s$ . Two two 68-kg linebackers attempt to stop him, one at -2.0 m/s and the other at -4.0 m/s. The negative value is because the run against the positive direction, taken in the direction of the fullback.

(a) Before the event, there is a total linear momentum, computed as the sum of the momentums of each player as shown

$p_1=m_1v_1=(98)(5)=490 Kg\ m/s$

$p_2=m_2v_2=(68)(-4)=-272 kg\ m/s$

$p_3=m_3v_3=(68)(-2)=-136 kg\ m/s$

$p_t=p_1+p_2+p_3=390-272-136=82\ kg\ m/s$

After the collision, all the players keep joined in one single mass of.

$m_t=98+68+68=234\ kg$

They will move at a speed which will be computed below

(b) The final momentum of the system is

$p_f=m_tv_f=82\ kg\ m/s$

Since the linear momentum is conserved, the final speed $v_f$ is common to all of the players. Let's solve to find it

$\displaystyle v_f=\frac{p_f}{m_t}$

$\displaystyle v_f=\frac{82}{234}$

$v_f=0.35\ m/s$

(c) Since the final speed of the players is positive, it means the touchdown was actually scored, the fullback moved forward across the goal line, the positive reference.

5 0

3 years ago

Example of the word médium ?

Ira Lisetskai [31]

I am the medium one that in the family

4 0

3 years ago

Brainlist nd 20 points

nadezda [96]

Answer:

I'm not sure but I think it's 35-39

4 0

2 years ago

Read 2 more answers

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

The output power is greater in the interval from 1.0 s to 2.5 s

Explanation:

Physical Power

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

3 years ago

How many minutes of daylight do we gain after winter solstice

serg [7]

Answer:

Depends on which hemisphere you are belong to and how much distance you are away from Ecuador.

Explanation:

Minutes of daylight is equal on everywhere only on the equinox days (21 March and 23 September). On other days it depends on the place that you are belong to. On winter solstice, places on Ecuador have 12 hours daylight. North side of Ecuador have less, south side of Ecuador have more hour of daylight.

4 0

3 years ago