PHYSICS COLLEGE

Michelle recently started selling her invention: A bed that looks like it floats in mid-air. The bed is actually suspended by magnetic forces. Michelle is a(n)

Answers

Answer 1

Answer:

Answer:

Explanation:

designer

illusionist

engineer

entrepreneur

salesperson

human

inventor

Related Questions

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snohomish River below. The swimmers stepped off the bridge, and I estimated that they hit the water 2.00 s later. A)How high was the bridge?B)How fast were the swimmers moving when they hit the water?C)What would the swimmer's drop time be if the bridge were twice as high?
A construction foreman exerts 1300 Newtons of force trying to move a 1200-kg block of concrete. How many Joules of work does he perform?
A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-sized mountain). Suppose an object falls from rest near the surface of such a star. How fast would this object be moving after it had fallen a distance of 0.025 m? (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)
Am example of a good electrical isulator isa.rubberb.ironc.copperd.aluminum
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

A 0.157kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.850m/s . It has a head-on collision with a 0.306kg glider that is moving to the left with a speed of 2.26m/s . Suppose the collision is elastic.Find the magnitude of the final velocity of the 0.157kg glider.Find the magnitude of the final velocity of the 0.306kg glider.

Answers

Answer:

V1 = -3.260 m/s, V2 = 1.303 m/s

Explanation:

Let mass of the left glider m1 = 0.157 kg and velocity v1 = 0.850 m/s

mass of the right glider m2 = 0.306 Kg and v2 = -2.26 m/s (-ve sign mean it is opposite to direction of left glider)

To Find: Final Velocity of Left Glider is V1=? m/s and Velocity of right Glider is V2 =? m/s (After Collision)

from law of conservation of momentum and energy we deduce a formula:

V1 = (m1-m2) v1 /(m1+m2) + 2 m2 v2/(m1+m2)

V1 = (0.157 kg - 0.306 Kg) × 0.850 m/s / (0.157 kg + 0.306 Kg) + 2 ×0.306 kg × -2.26 m/s / (0.157 kg + 0.306 Kg)

V1 = -0.273 -2.987

V1 = -3.260 m/s

and V2 Formula

V2 = (m2-m1) v2/(m1+m2) + 2 m1 v1/(m1+m2)

V2 = (0.157 kg - 0.306 Kg) × -2.26 m/s / (0.157 kg + 0.306 Kg) + 2 ×0.157 kg × 0.850 m/s / (0.157 kg + 0.306 Kg)

V2 = 0.727 + 0.576

V2 = 1.303 m/s

-0.149, 0.463

How many meters will a car travel if its speed is 45 m/s in an interval of 11 seconds?Question 2 options:

A) 450 meters

B) 495 meters

C) 4.09 meters

D) 498 meters

Answers

Data given:

V=45m/s

t=11s

Δx=?

Formula needed:

V=Δx/t

Solution:

Δx=v×t

Δx=45m/s×11s

Δx=495m

According to my solution B) is the most accurate

495 is the answer !!

Why is wood a renewable resource?A) We use it over and over again.
B) It takes millions of years to replace.
C) Trees grow everywhere, so we always have some.
D) It can be replaced in a reasonable amount of time.

Answers

Answer:

D) It can be replaced in a reasonable amount of time.

Explanation:

i took the test!!!!!1

An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless spring, compressing it a maximumdistance of 0.750m. (See below.) The spring constant is 500N/m, the height of the incline is 2.0 m, and the horizontalsurface is frictionless. (a) What is the speed of the object atthe bottom of the incline? (b) What is the work of frictionon the object while it is on the incline? (c) The springrecoils and sends the object back toward the incline. Whatis the speed of the object when it reaches the base of theincline? (d) What vertical distance does it move back up theincline?

Answers

Final answer:

The final speed at the bottom of the incline can be calculated using the conservation of energy principle. There is no work done against friction as the object is moving on a frictionless surface. The speed does not change when the spring pushes it back towards the base of the incline due to lack of friction and it moves to a certain height given the angle of the incline and the initial speed.

Explanation:

"Speed at the bottom of the incline:" This can be calculated using conservation of energy. The potential energy at the top (m*g*h) will convert into kinetic energy at the bottom (1/2*m*v^2). Here, m is the mass, g is acceleration due to gravity, h is the height, and v is the velocity. Using this, we can solve for v.
Work of friction on the incline: As per the question, the surface is frictionless. Therefore, the work done by friction is automatically 0 as there is no force of friction.
Speed of the object when it reaches the base of the incline again: As the surface is frictionless, the object reaches the incline with the same speed with which it left as there are no opposing forces to reduce its momentum.
Vertical distance it moves back up the incline: This can be calculated using the principles of conservation of energy and kinematic equations, taking into account the angle of the incline and the velocity of the object.

Learn more about Mechanics and Energy Conservation here:

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A long wire carries a current density proportional to the distance from its center, J=(Jo/ro)•r, where Jo and ro are constants appropriate units. Determine the magnetic field vector inside this wire.

Answers

Answer:

$B = \mu_0((1)/(3) (J_0)/(r_0) r^2)$

Explanation:

As the current density is given as

$J = (J_0)/(r_0)r$

now we have current inside wire given as

$i = \int J(2\pi r)dr$

$i = \int (J_0)/(r_0) r(2\pi r)dr$

$i = 2\pi (J_0)/(r_0) \int r^2 dr$

$i = (2)/(3) \pi (J_0)/(r_0) r^3$

Now by Ampere's law we will have

$\int B. dl = \mu_0 i$

$B. (2\pi r) = \mu_0((2)/(3) \pi (J_0)/(r_0) r^3)$

$B = \mu_0((1)/(3) (J_0)/(r_0) r^2)$

Proposed Exercise - Circular MovementConsider four pulleys connected by correals as illustrated in the figure below. One motor moves the A pulley with angular acceleration A= 20 rad/s^2/. If pulley A is initially moving with angular acceleration A= 40 rad/s^2, determine the angular speed of pulleys B and C after three seconds. Consider that the belts do not slide

Answers

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

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A centrifuge is a common laboratory instrument that separates components of differing densities in solution. This is accomplished by spinning a sample around in a circle with a large angular speed. Suppose that after a centrifuge in a medical laboratory is turned off, it continues to rotate with a constant angular deceleration for 10.0s before coming to rest.Part AIf its initial angular speed was 3890rpm , what is the magnitude of its angular deceleration? (|?| in revs/s^2 )Part BHow many revolutions did the centrifuge complete after being turned off?

A train accelerates at -1.5 m/s2 for 10 seconds. If the train had an initialspeed of 32 m/s, what is its new speed?A. 17 m/sB. 15 m/sC. 47 m/sD. 32 m/s

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