A bucket of water with weight of 50N is pulled up using a rope. The bucket is moving at

valentina_108 [34]

<span>1 C = 6.24150965(16)×10^18 electrons

31.25 x 10^18 electrons / (6.24150965(16)×10^18 electrons / C) = 5.007 Coulombs

</span><span>I hope this helps. </span>

4 0

2 years ago

Read 2 more answers

What is the value of the equivalent resistance for the three resistors connected in series?

Harman [31]

The value of the equivalent resistance for the three resistors connected in series will be the sum of the three values.

To find the answer, we have to know more about the equivalent resistance.

<h3>What is meant by equivalent resistance?</h3>

equivalent resistance is the total value of the resistance connected in a circuit.
If n resistors are connected in series, then the equivalent resistance will be,

$R_E=R_1+R_2+..........+R_n$

In our question we have three resistors. Thus, the equivalent resistance will be,

$R_E=R_1+R_2+R_3$

Thus, we can conclude that, the value of the equivalent resistance for the three resistors connected in series will be the sum of the three values.

Learn more about the equivalent resistance here:

brainly.com/question/11603204

#SPJ4

7 0

1 year ago

A third baseman makes a throw to first base 40.5 m away. The ball leaves his hand with a speed of 30.0 m/s at a height of 1.4 m

Ugo [173]

Answer:

When the ball goes to first base it will be 4.23 m high.

Explanation:

Horizontal velocity = 30 cos17.3 = 28.64 m/s

Horizontal displacement = 40.5 m

Time

$t=\frac{40.5}{28.64}=1.41s$

Time to reach the goal posts 40.5 m away = 1.41 seconds

Vertical velocity = 30 sin17.3 = 8.92 m/s

Time to reach the goal posts 40.5 m away = 1.41 seconds

Acceleration = -9.81m/s²

Substituting in s = ut + 0.5at²

s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m

Height of throw = 1.4 m

Height traveled by ball = 2.83 m

Total height = 2.83 + 1.4 = 4.23 m

When the ball goes to first base it will be 4.23 m high.

8 0

3 years ago

SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the

Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0

2 years ago

A 60 W lightbulb is on for 24 hours. A stereo with a power of 150 W is on for 2

Helen [10]

Answer:

B) The lightbulb uses 4,104,000 J more than the stereo.

6 0

2 years ago