Go to his profile and roast the mess out of him plzz 403665fl 50 points

Answers

Answer 1
Answer:

Answer:

ok

Explanation:


Related Questions

A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75. How many electrons are delivered to the phone during 27 min of charging?
Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stock on hand as acquired over the years. Readily available are two common resistorsrated at 500±50 Ωand two common resistors rated at 2000 Ω±5%. What isthe uncertainty in an equivalent 400 Ωresistance?(Hint: the equivalent resistance connected in parallel can be obtained by 1212TRRRRR=+)
What is the force required to accelerate 1 kilogram of mass at 1 meter per second per second.1 Newton 1 pound 1 kilometer 1 gram
Part A (4 pts) Consider light of wavelength λ = 670nm traveling in air. The light is incident at normal incidence upon a thin film of oil with n2 =1.75. On the other side of the thin film is glass with n3 =1.5. What is the minimum non-zero value of the film thickness d that will cause the reflections from both sides of the film to interfere constructively?
Find the magnitude and direction of an electric field that exerts a 4.80×10−17N westward force on an electron. (b) What magnitude and direction force does this field exert on a proton?

A ride at an amusement park moves the riders in a circle at a rate of 6.0 m/s. If the radius of the ride is 9.0 meters, what is the acceleration of the ride?4.0 m/s2
0.67 m/s2
0.075 m/s2
54 m/s2

Answers

4.0 m/s2


it's 9 squared divided by 6
the answer is B 4.0 m/s2

A farmer is using a rope and pulley to lift a bucket of water from the bottom of a well. the farmer uses a force f1=57.5 n to pull the bucket of water upwards. the total mass of the bucket of water is f2= 3.9kg. -Calculate how much work Wg in J gravity does on the bucket filled with water as the farmer lifts it up the well.
-Calculate the net work Wnet in J done on the bucket of water by the two forces F1 and Fg.

Answers

Final answer:

To calculate the work done by gravity on the bucket of water as it is lifted up the well, multiply the weight of the bucket by the lifting distance. The net work done on the bucket by the force applied by the farmer and gravity is the sum of the work done by both forces. The net work is represented by the equation Wnet = W1 + Wg.

Explanation:

To calculate how much work gravity does on the bucket filled with water as the farmer lifts it up the well, we need to multiply the force of gravity (weight) by the vertical distance the bucket is lifted. The equation for work is W = Fd, where W is the work done, F is the force, and d is the distance. In this case, the force of gravity is the weight of the bucket, which can be calculated by multiplying the mass by the acceleration due to gravity (9.8 m/s^2).

So, the work done by gravity (Wg) on the bucket is Wg = Fg * d = (m * g) * d = (3.9 kg * 9.8 m/s^2) * d = 38.22 d Joules.

To calculate the net work done on the bucket by the two forces, we can use the equation Wnet = W1 + Wg, where W1 is the work done by force F1 and Wg is the work done by gravity. Since force F1 and the displacement (lifting distance) are both vertical, the work done by F1 is given by W1 = F1 * d.

Therefore, the net work done on the bucket by forces F1 and gravity is Wnet = F1 * d + Fg * d = (57.5 N) * d + (3.9 kg * 9.8 m/s^2) * d = (57.5 N + 38.22 d) Joules.

Learn more about Work done by forces here:

brainly.com/question/31428590

#SPJ12

A tightly wound solenoid is 15 cm long, has 350 turns, and carries a current of 4.0 A. If you ignore end effects, you will find that the value of app at the center of the solenoid when there is no core is approximately

Answers

Answer:

The magnetic field at the center of the solenoid is approximately  0.0117 T

Explanation:

Given;

length of the solenoid, L = 15 cm = 0.15 m

number of turns of the solenoid, N = 350 turns

current in the solenoid, I = 4.0 A

The magnetic field at the center of the solenoid is given by;

B = \mu_o ((N)/(L) )I\n\nB = (4 \pi *10^(-7))((350)/(0.15) )(4.0)\n\nB = 0.0117 \ T

Therefore, the magnetic field at the center of the solenoid is approximately  0.0117 T.

24-gauge copper wire has a diameter of 0.51 mm. The speaker is located exactly 4.27 m away from the amplifier. What is the minimum resistance of the connecting speaker wire at 20°C? Hint: How many wires are required to connect a speaker!Compare the resistance of the wire to the resistance of the speaker (Rsp = 8 capital omega)

Answers

Answer:

 R = 8.94 10⁻² Ω/m,    R_sp / R_total = 44.8

Explanation:

The resistance of a metal cable is

         R = ρ L / A

The area of ​​a circle is

          A = π R²

The resistivity of copper is

        ρ = 1.71 10⁻⁸ ohm / m

Let's calculate

       R = 1.71 10⁻⁸  4.27 / (π (0.51 10⁻³)²)

       R = 8.94 10⁻² Ω/m

Each bugle needs two wire, phase and ground

The total wire resistance is

        R_total = 2 R

        R_total = 17.87 10⁻² Ω

Let's look for the relationship between the resistance of the bugle and the wire

      R_sp / R_total = 8 / 17.87 10⁻²

      R_sp / R_total = 44.8

Final answer:

The resistance of the speaker wire can be calculated using the formula for the resistance of a wire, taking into account the resistivity of copper, the length and thickness of the wire, and whether a single or pair of wires is used.

Explanation:

The question is asking you to find the minimum resistance of a copper wire given its diameter and length, plus the resistance of the speaker it's connected to. Resistance of a wire is calculated using the formula R=ρL/A, where R is the resistance, ρ (rho) is the resistivity of the material (in this case, copper), L is the length of the wire, and A is the cross-sectional area of the wire.

First, you need to find the area of the 0.51 mm diameter wire. The area (A) of a wire is given by the formula π(d/2)^2 where d is the diameter of the wire. After calculating the area, use the formula R=ρL/A to calculate the resistance. For copper wire at 20°C, ρ is approximately 1.68 × 10^-8 Ω·m. Substituting these values into the formula will give you the resistance of the wire in ohms.

Note: you may need to consider whether you have just a single wire or a pair, since two wires are typically required to connect a speaker. If a pair is used, each wire will carry half the current, which affects the total resistance.

Learn more about Electric Resistance here:

brainly.com/question/31668005

#SPJ12

You have a grindstone (a disk) that is 105.00 kg, has a 0.297-m radius, and is turning at 71.150 rpm, and you press a steel axe against it with a radial force of 46.650 N. Assuming the kinetic coefficient of friction between steel and stone is 0.451. How many turns will the stone make before coming to rest?

Answers

Answer:

3.27 turns

Explanation:

To find how many turns (θ) will the stone make before coming to rest we will use the following equation:

\omega_(f)^(2) = \omega_(0)^(2) + 2\alpha*\theta

Where:

\omega_(f): is the final angular velocity = 0

\omega_(0): is the initial angular velocity = 71.150 rpm

α: is the angular acceleration

First, we need to calculate the angular acceleration (α). To do that, we can use the following equation:

\alpha = (\tau)/(I)

Where:

I: is the moment of inertia for the disk

τ: is the torque  

The moment of inertia is:

I = (mr^(2))/(2)

Where:

m: is the mass of the disk = 105.00 kg

r: is the radius of the disk = 0.297 m

I = (105.00 kg*(0.297 m)^(2))/(2) = 4.63 kg*m^(2)

Now, the torque is equal to:

\tau = -F x r = -\mu*F*r

Where:

F: is the applied force = 46.650 N      

μ: is the kinetic coefficient of friction = 0.451

\tau = -\mu*F*r = -0.451*46.650 N*0.297 m = -6.25 N*m

The minus sign is because the friction force is acting opposite to motion of grindstone.    

Having the moment of inertia and the torque, we can find the angular acceleration:

\alpha = (-6.25 N*m)/(4.63 kg*m^(2)) = -1.35 rad/s^(2)

Finally, we can find the number of turns that the stone will make before coming to rest:

0 = \omega_(0)^(2) + 2\alpha*\theta  

\theta = -((\omega_(0))^(2))/(2\alpha) = -((71.150 (rev)/(min))^(2))/(2*(-1.35 (rad)/(s^(2)))*(1 rev)/(2\pi rad)*((60 s)^(2))/((1 min)^(2))) = 3.27 rev = 3.27 turns                        

I hope it helps you!

A 5.49 kg ball is attached to the top of a vertical pole with a 2.15 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.654.65 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take ????=9.81g=9.81 m/s2. angle: °

Answers

Answer:\theta =45.73^(\circ)

Explanation:

Given

Length of string =2.15 m

mass of ball =5.49 kg

speed of ball=4.65 m/s

Here

Tension provides centripetal acceleration

T\cos\theta =mg-----1

T\sin \theta =(mv^2)/(r)------2

Divide 2 & 1

tan\theta =(v^2)/(rg)

tan\theta =(4.65^2)/(2.15* 9.8)

tan\theta =1.026

\theta =45.73^(\circ)