An athlete can exercise by making mechanical waves in ropes. What is themedium of these waves?
A. Energy
B. The rope
C. The athlete
D. Air

Answers

Answer 1

Answer:

Answer:

It would be B. The Rope

Explanation:

I say this because the rope is transferring energy from one location to another. Now, I could be totally wrong on this but I think this is right lol.

Answer 2

Answer:

Answer:

The answer is B the rope.

Hope this helps <3 ;)

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The index of refraction for red light in water is 1.331, and that for blue light is 1.340. If a ray of white light enters the water at an angle of incidence of 83.00o , what are the underwater angles of refraction for the blue and red components of the light
You have built a device that measures the temperature outside and displays it on a dial as a measure of how far away from room temperature outside is. The way the dial works is that a needle with a charged ball on the end is placed between two charged parallel plates. The strength of the uniform electric field between the plates is proportional to the outside temperature. Given that the charged ball on the needle has a charge of?
Suppose that an object undergoes simple harmonic motion, and its displacement has an amplitude A = 15.0 cm and a frequency f = 11.0 cycles/s (Hz). What is the maximum speed ( v ) of the object?A. 165 m/sB. 1.65 m/sC. 10.4 m/sD. 1040 m/s
Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.a. in what direction, relative to the shore, does dave’s boat go? b. how long does it take dave to cross the river? c. how far downstream is dave’s landing point? d. how long would it take dave to cross the river if there were no current?

High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of the track depends on the speed of the particle and its lifetime. A particle moving at 0.993c leaves a track 1.15 mm long. What is the proper lifetime of the particle

Answers

Answer:

Lifetime = 4.928 x 10^-32 s

Explanation:

(1 / v2 – 1 / c2) x2 = T2

T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225

T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6

T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s

Final answer:

To find the proper lifetime of the particle, we can use the time dilation equation and the Lorentz factor. Plugging in the given values, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.

Explanation:

To find the proper lifetime of the particle, we can use the time dilation equation, which states that the proper time (time experienced in the frame of reference of the particle) is equal to the time observed in the laboratory frame of reference divided by the Lorentz factor. The Lorentz factor can be calculated using the equation γ = 1/√(1 - (v/c)^2), where v is the velocity of the particle and c is the speed of light. Given that the particle is moving at 0.993c, the Lorentz factor is approximately 22.82.

Next, we can use the equation Δx = βγcτ, where Δx is the length of the track, β is the velocity of the particle in units of the speed of light (v/c), γ is the Lorentz factor, c is the speed of light, and τ is the proper lifetime of the particle. Plugging in the given values, we have 1.15 mm = 0.993 * 22.82 * c * τ. Solving for τ, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.

Learn more about Proper lifetime of high-energy particles here:

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In an evironmental system of subsystem, the mass balance equation is:__________.

Answers

Answer:

Explanation:

The mass balance is an application of conservation of mass, to the analysis of physical system. This is given in an equation form as

Input = Output + Accumulation

The conservation law that is used in this analysis of the system actually depends on the context of the problem. Nevertheless, they all revolve around conservation of mass. By conservation of mass, I mean that the fact that matter cannot disappear or be created spontaneously.

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?

Answers

Explanation:

Given

Acceleration of the pebble is

At t=0, velocity is

considering horizontal motion

$\Rightarrow x=ut+0.5at^2 \n\Rightarrow 11=4.3t+0.5(4.6)t^2\n\Rightarrow 2.3t^2+4.3t-11=0\n\Rightarrow (t-1.4435)(t+3.3131)=0\n\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\n$

Velocity acquired during this time

$\Rightarrow v_x=4.3+4.6* 1.44\n\Rightarrow v_x=4.3+6.624\n\Rightarrow v_x=10.92\ s$

Consider vertical motion

$\Rightarrow v_y=0+7(1.44)\n\Rightarrow v_y=10.08\ m/s$

Net velocity is

$\Rightarrow v=√(10.92^2+10.08^2)\n\Rightarrow v=√(220.85)\n\Rightarrow v=14.86\ m/s$

Angle made is

$\Rightarrow \tan \theta =(10.08)/(10.92)\n\n\Rightarrow \tan \theta =0.92307\n\n\Rightarrow \theta =42.7^(\circ)$

QuestIuI(2 PUMILS)
How much power is needed to lift a 750 kg elephant 14.3 m in 30.0 seconds?

Answers

Given Information:

Mass of elephant = m = 750 kg

Height = h = 14.3 m

time = t = 30 seconds

Required Information:

Power needed to lift elephant = P = ?

Answer:

Power needed to lift elephant ≈ 3507 watts

Explanation:

As we know power is given by

P = PE/t

Where PE is the potential energy and t is the time

Potential energy is given by

PE = mgh

Where m is the mass of elephant, g is the gravitational acceleration and h is the height to lift the elephant.

PE = 750*9.81*14.3

PE = 105212.25 Joules

Therefore, the required power to lift the elephant is

P = PE/t

P = 105212.25/30

P ≈ 3507 watts

Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by the discharge of a 1.89 Fcapacitor charged to 60.9 kV. Rosa rightly reckons that she can enhance the effect of each laser pulse by increasing the electric potential energy of the charged capacitor. She could do this by replacing the capacitor's filling, whose dielectric constant is 431, with one possessing a dielectric constant of 947.Required:
a. Find the electric potential energy of the original capacitor when it is charged. (in Joules)
b. Calculate the electric potential energy of the upgraded capacitor when it is charged. ( In Joules)

Answers

Answer:

$U = 3.505 *10^9 \ J$

$U_1 = 7.696 *10^9 \ J$

Explanation:

From the question we are told that

The capacitance is $C = 1.89 \ F$

The voltage is $V = 60.9 \ k V = 60.9 *10^(3) \ V$

The first dielectric constant is $\epsilon_1 = 431$

The second dielectric constant is $\epsilon_2 = 947$

Generally the electric potential energy is mathematically represented as

$U = (1)/(2) * C * V^2$

=> $U = (1)/(2) * 1.89 * (60.9 *10^(3))^2$

=> $U = 3.505 *10^9 \ J$

Generally the capacitance when the capacitor's filling was changed is

$C_n = 1.89 * (947)/(431)$

=> $C_n = 4.15$

Generally the electric potential energy when the capacitor's filling was changed is

$U_1 = (1)/(2) * C_1 * V^2$

=> $U_1 = (1)/(2) * 4.15 * (60.9 *10^(3))^2$

=> $U_1 = 7.696 *10^9 \ J$

An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?

Answers

Answer:

11405Volt

Explanation:

To solve this problem it is necessary to use the concept related to induced voltage or electromotive force measured in volts. Through this force it is possible to maintain a potential difference between two points in an open circuit or to produce an electric current in a closed circuit.

The equation that allows the calculation of this voltage is given by,

$\epsilon = BAN \omega$