If two teams playing tug-of-war pull on a rope with equal but opposite forces, what is the net external force on the rope?

What is an emergent ray

dangina [55]

Answer:

Explanation:

the light ray leaving a medium in contrast to the entering or incident ray.

5 0

2 years ago

A boy throws a ball and accidentally breaks a window. The momentum of the ball and all the pieces of glass taken together after

geniusboy [140]

A boy throws a ball and accidentally breaks a window. The momentum of the ball and all the pieces of glass taken together after the collision is the same as <span>the momentum of the ball before the collision. I think you forgot to give the choices along with the question. I hope that the answer has come to your great help.</span>

3 0

3 years ago

A boy throws a ball with an initial velocity of 19.6 m/s. What maximum height does the ball reach?

Wewaii [24]

<h2>Hello!</h2>

The answer is: 19.59 m

Since there is no information about the launch type, we can assume that the ball is thrown vertically upward.

When the ball reaches the maximum height, just at that moment, the velocity turns to 0, and after that moment, the ball starts falling, so:

We will use the following formula:

$Vf^2=Vi^2+2*g*s$

Where:

Vf= Final velocity = 0

Vi= Initial velocity = $\frac{19.6m}{s}$

g = Gravity Acceleration = $\frac{9.81m}{s^{2} }$

s = Traveled distance

$0=19.6^2+2*-9.81*s\\s=\frac{19.6^2}{2*9.81}=\frac{384.16}{19.62}=19.59m$

Have a nice day!

8 0

3 years ago

The voltage across the terminals of a 250nF capacitor is푣푣=�50푉푉, 푡푡≤0(푚푚1푒푒−4000푡푡+푚푚2푡푡푒푒−4000푡푡)푉푉, 푡푡 ≥0The initial current

olga2289 [7]

The first part of the question is not complete and it is;

The voltage across the terminals of a 250 nF capacitor is 50 V, A1e^(-4000t) + (A2)te^(-4000t) V, t0, What is the initial energy stored in the capacitor? Express your answer to three significant figures and include the appropriate units. t

Answer:

A) initial energy = 0.3125 mJ

B) A1 = 50 and A2 = 1,800,000

C) Capacitor Current is given by the expression;

I = e^(-4000t)[0.95 - 1800t]

Explanation:

A) In capacitors, Energy stored is given as;

U = (1/2)Cv²

Where C is capacitance and v is voltage.

So initial kinetic energy;

U(0) = (1/2)C(vo)²

From the question, C = 250 nF and v = 50V

So, U(0) = (1/2)(250 x 10^(-9))(50²) = 0.3125 x 10^(-3)J = 0.3125 mJ

B) from the question, we know that;

A1e^(-4000t) + (A2)te^(-4000t)

So, v(0) = A1e^(0) + A2(0)e^(0)

v(0) = 50

Thus;

50 = A1

Now for A2; let's differentiate the equation A1e^(-4000t) + (A2)te^(-4000t) ;

And so;

dv/dt = -4000A1e^(-4000t) + A2[e^(-4000t) - 4000e^(-4000t)

Simplifying this, we obtain;

dv/dt = e^(-4000t)[-4000A1 + A2 - 4000A2]

Current (I) = C(dv/dt)

I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]

Thus, Initial current (Io) is;

Io = (250 x 10^(-9))[e^(0)[-4000A1 + A2]]

We know that Io = 400mA from the question or 0.4 A

Thus;

0.4 = (250 x 10^(-9))[-4000A1 + A2]

0.4 = 0.001A1 - (250 x 10^(-9)A2)

Substituting the value of A1 = 50V;

0.4 = 0.001(50) - (250 x 10^(-9)A2)

0.4 = 0.05 - (250 x 10^(-9)A2)

Thus, making A2 the subject, we obtain;

(0.4 + 0.05)/(250 x 10^(-9))= A2

A2 = 1,800,000

C) We have derived that ;

I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]

So putting values of A1 = 50 and A2 = 1,800,000 we obtain;

I = (250 x 10^(-9))e^(-4000t)[(-4000 x 50) + 1,800,000 - 4000(1,800,000)t]

I = e^(-4000t)[0.05 + 0.45 - 1800t]

I = e^(-4000t)[0.95 - 1800t]

5 0

4 years ago

Which part of a stream's sediment load moves the slowest?

Rudiy27

The bed load moves the slowest from all the parts of the stream's sediment. It consists of particles suspended that are suspended and float around the bed. This part is the slowest in motion, as it rolls, and moves with the flow. The particles near the bed are not dissolved so they settle at the bottom and move with the stream.

4 0

3 years ago