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Ira Lisetskai [31]

2 years ago

5

Block A has mass 1.00 kg and block B has mass 3.00 kg. The blocks collide and stick together on a level, frictionless surface. A

fter the collision, the kinetic energy (KE) of block A is

1 answer:

Pani-rosa [81]2 years ago

5 0

Answer:

1/2mv²=0

1/2(4kg)(v²)=0

2=-v²

square root -2=v

v=1.414

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Which animal in the rainforest is not nocturnal?

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A 1200 kg elevator accelerated upwards at 2 m/s2. Draw a force diagram for the elevator. Calculate the force of tension in the c

pishuonlain [190]

Answer:

4800

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3 years ago

A point charge +2Q is at the origin and a point charge −Q is located along the x axis at x = d as in the figure below. Find a sy

Akimi4 [234]

Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis

$F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}$

Explanation:

Let the force on +Q charge y-axis due to +2Q charge be $F_1$ and force on +Q charge y axis due to -Q charge on x-axis be $F_2$ .

Distance between the +2Q charge and +Q charge = d units

Distance between the -Q charge and +Q charge = $\sqrt{2}d$ units

$k_e$ = Coulomb constant

$F_1=k_e\frac{(+2Q)(+Q)}{d^2}=k_e\frac{+2Q^2}{d^2} N$

$F_2=k_e\frac{(-Q)(+Q)}{(\sqrt{2}d)^2}=k_e\frac{-Q^2}{2d^2} N$

Net force on +Q charge on y-axis is:

$F_x=F_2sin 45^o=k_e\frac{-Q^2}{2d^2}\times \frac{1}{\sqrt{2}} N$

$F_y=F_1-F_2cos45^o$

$F_y=(F_1-F_2cos45^o)=(k_e\frac{+2Q^2}{d^2})-(k_e\frac{-Q^2}{2d^2}\frac{1}{\sqrt{2}})$

$F_N=\sqrt{F_x^2+F_y^2}$

$|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|$

The net froce on the +Q charge on y-axis is

$F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}$

4 0

3 years ago

Read 2 more answers

a circular loop is hanging on the wall. it has a radius of 33.3 cm and is comprised of 12 coils. there is a magnetic field perpe

Vika [28.1K]

Answer:

I = 2.19A, anticlockwise direction.

Explanation:

Given r = 33cm = 0.33m, N = 12, ΔB = 7.5 - 1.5 = 6.0T, Δt = 3s, R = 3.75Ω

By Faraday's law of electromagnetic induction when there is a change in flux in a coil or loop, an emf is induced in the coil or loop which is proportional to the time rate of change of the magnetic flux through the loop.

The emf E is related to the flux by the formula

E = – NdФ/dt

Where N = number of turns in the coil, Ф = magnetic flux through the loop = BA, B = magnetic field strength, A = Area

In this problem the strength of the magnetic field changes. As a result the flux too changes and an emf is induced in the coil.

So

ΔФ = ΔB×A = ΔB×πr² = 6×π×0.33² = 2.05Wb

E = -NΔФ/Δt = 12×2.05/3 = 8.2V

I = E/R = 8.2/ 3.75 = 2.19A

The direction of the current can be found by pointing the thumb of your right hand in the direction of the magnetic field and curling the remaining fingers around this direction. The direction of the curl of these fingers give the direction of current which in this case is anticlockwise.

5 0

3 years ago