PHYSICS COLLEGE

The data listed below are for mechanical waves. Which wave has the greatest energy?

Answers

Answer 1

Answer: theres no data listed below

Answer 2

Answer:

Answer:

amplitude = 14 cm; wavelength = 7 cm; period = 12 seconds

Explanation:

Related Questions

M = 30.3kgM = 40.17kg 9R = 0.5mG = 6. 67x10^11F ?
The height of the upper falls at Yellowstone Falls is 33 m. When the water reaches the bottom of the falls, its speed is 26 m/s. Neglecting air resistance, what is the speed of the water at the top of the falls?
A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to move it 4.00 m to the side.(a) What is the magnitude of F when the crate is in this final position? During the crate's displacement, what are (b) the total work done on it, (c) the work done by the gravitational force on the crate, and (d) the work done by the pull on the crate from the rope? (e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force F does on the crate. (f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?
Two bullets of the same size, mass and horizontal velocity are fired at identical blocks, only one is made of steel and the other is made of rubber. The steel bullet has a perfectly inelastic collision with the block, while the rubber bullet has an elastic collision. Which bullet is more likely to knock over the block, or are both equally likely to do so? Justify your choice based on physics principles.
The acceleration of a particle moving along a straight line is given by a = −kt2 m/s2 where k is a constant and time t is in seconds. The initial velocity of the particle at t = 0 is v0 = 12 m/s and the particle reverses it direction of motion at t = 6 s. Determine the constant k and the displacement of the particle over the same 6-second interval of motion. Ans: k = 1/6 m/s4, Δs = 54 m

What is the weight of a 5.0-kilogram object at the surface of Earth?A
5.0 kg
B
25 N
C
49 kg
D
49 N

Answers

The weight of a 5.0-kilogram object at the surface of Earth is 49 N. Hence, option (C) is correct.

What are mass and weight?

Mass is a fundamental quantity in physics and the most fundamental attribute of matter. Mass can be defined as the amount of matter contained in a body. Kilogram is the SI unit of mass (kg).

A body's mass does not alter at any time. Only in severe instances where a massive amount of energy is provided or removed from the body.

The force of gravity acting on a body is measured by weight. The weight formula is as follows: w = mg. Because weight is a force, it has the same SI unit as force; the SI unit of weight is Newton (N).

The weight of the object is = 5.0 × 9.8 N

= 49.0 N.

Hence, option (B) is correct.

Learn more about weight here:

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Answer:

49 N (d)

Explanation:

w= mg = 5 kg * 9.8 m/s^2 = 49 N

What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.

Answers

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

On a 30degrees day, there is an explosion. The sound is heard 3.4s after seeing the flash. How far away was the explosion

Answers

Answer:

Distance = 1.2 km (Approx)

Explanation:

Given:

Temperature (T) = 30°C

Total Time taken (t) = 3.4 Sec

We know that sound increases in the air sound increase nearly 0.60 m/s for every sound with temperature.

Speed of sound = 331 m/s

So,

V = (331 + 0.60T) m/s

V = 331+(0.60×300C) m/s

V = 349 m/s

Distance = speed × time

Distance = 349 × 3.4

Distance = 1,186.6 m

Distance = 1.2 km (Approx)

In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)

Answers

Answer:

v(t) = 21.3t

v(t) = 5.3t

$v(t) = 48 -48 e ^{ (t)/(9)}$

Explanation:

When no sliding friction and no air resistance occurs:

$m(dv)/(dt) = mgsin \theta$

where;

$(dv)/(dt) = gsin \theta , 0 < \theta < ( \pi)/(2)$

Taking m = 3 ; the differential equation is:

$3 (dv)/(dt)= 128*(1)/(2)$

$3 (dv)/(dt)= 64$

$(dv)/(dt)= 21.3$

By Integration;

$v(t) = 21.3 t + C$

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta$

Taking m =3 ; the differential equation is;

$3 (dv)/(dt)=128*(1)/(2) -(√(3) )/(4)*128*(√(3) )/(4)$

$(dv)/(dt)= 5.3$

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta - kv$

The differential equation is :

= $3 (dv)/(dt)=128*(1)/(2) - ( √( 3))/(4)*128 *( √( 3))/(2)-(1)/(3)v$

= $3 (dv)/(dt)=16 -(1)/(3)v$

By integration

$v(t) = 48 + Ce ^{(t)/(9)$

Since; V(0) = 0 ; Then C = -48

$v(t) = 48 -48 e ^{ (t)/(9)}$

A rectangular loop (area = 0.15 m2) turns in a uniform magnetic field, B = 0.18 T. When the angle between the field and the normal to the plane of the loop is π/2 rad and increasing at 0.75 rad/s, what emf is induced in the loop?

Answers

Answer:

Emf induced in the loop is 0.02V

Explanation:

To get the emf of induced loop, we have to use faraday's law

ε = - dΦ/dt

To get the flux, we use;

Φ = BA cos(θ)

B = The uniform magnetic field

A = Area of rectangular loop

θ = angle between magnetic field and normal to the plane of loop

substitute the flux equation (Φ) into the faraday's equation

we have ε = - d(BA cos(θ)) / dt

ε = BA sinθ dθ/dt

from the question;B = 0.18T, A=0.15m2, θ = π/2 ,dθ/dt = 0.75rad/s

Our equation will now look like this;

ε = (0.18T) (0.15m2) (sin(π/2)) (0.75rad/s)

ε = 0.02V

While testing at 30 feet below the surface in Lake Minnetonka, with the sub stopped and in equilibrium, one of the students aboard the sub drops a hammer that goes through the hull of the submarine, and sticks out of the submarine handle first. When this happens, a seal forms immediately around the handle, so that no water enters the sub. What is the new equilibrium position for the sub?

Answers

Answer:

Explanation:

The equilibrium position of the sub is at the surface of the lake

Other Questions

Two long ideal solenoids (with radii 20 mm and 30 mm, respectively) have the same number of turns of wire per unit length. The solenoid is mounted inside the larger, along a common axis. The magnetic field with in the inner solenoid is zero. The current in the inner solenoid must be: a. two-thirds the current in the outer solenoid b. one-third the current in the outer solenoid c. twice the current in the outer solenoid d. half of the current in the outer solenoid e. the same as the current in the outer solenoid

In an 8.00km race, one runner runs at a steady 11.8 km/hr and another runs at 15.0 km/hr. How far from the finish line is the slower runner when the faster runner finishes the race?

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

Your classmate’s mass is 63 kg and the table weighs 500 N. Calculate the normal force on the table by the floor. Show your work!