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klemol [59]
2 years ago
13

A small box of mass m1 is sitting on a board of mass m2 and length L (Figure 1) . The board rests on a frictionless horizontal s

urface. The coefficient of static friction between the board and the box is ?s. The coefficient of kinetic friction between the board and the box is, as usual, less than ?s.
Throughout the problem, use g for the magnitude of the acceleration due to gravity. In the hints, use Ff for the magnitude of the friction force between the board and the box.

Find Fmin, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).

Express your answer in terms of some or all of the variables ?s, m1, m2, g, and L. Do not include Ff in your answer.
Physics
1 answer:
chubhunter [2.5K]2 years ago
3 0

Explanation:

Whole system will accelerate under the action of applied force. The box will experience the force against the friction and when this force exceeds then the box will move. so

Ff = μs×m1×g

m1×a = μs×m1×g

a = μs×g

The applied force is given by

F = (m1 + m2)×a so

F = μs×g×(m1+m2)

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A simple generator has a square armature 6.0 cm on a side. The armature has 85 turns of 0.59-mm-diameter copper wire and rotates
FrozenT [24]

Answer:

f=15.5 Hz

Explanation:

Let's determine the internal resistance:

R=\frac{(p*L)}{A}

ρ = 1.68*10^-8 Ω m

L=0.060m*4*60 = 14.4m

A=\pi*r^2\\A= \pi*(5.9x10^-4m/2)^2=2.734*10^-7m^2

R=(1.68*10^-8)*(14.4m)/(2.734*10^-7m^2)= 0.884Ω

Since the bulb is rated at 12.0 V and 25.0 W,

Current

I=\frac{25W}{12.0v}=2.08 A

Therefore, voltage drop inside generator =

V=(2.08 A)*(0.88)=2.35v

Actual EMF required is

E_{mf}=12.0v+2.35v=14.35v

Note that this is an RMS value.  

The peak voltage is

v_{peak}=14.15v*\sqrt{2} =20.29v

For a generator, by Faraday's Law,

E_{(max)}=N*B*A*w

20.29v=(60)*(0.650T)*(0.06m)^2*ω

ω=144.5\frac{rad}{s}

f=ω/(2π)=

f=144.5 rad/s/(2π)

f=23.001 Hz

6 0
2 years ago
A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?
OLga [1]

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

7 0
2 years ago
Read 2 more answers
A person in a boat sees a fish in the water (n = 1.33, the light rays making an angle of 40? relative to the water's surface. wh
jek_recluse [69]
We know that the measure of an incident ray is:  α 1 = 40°.
The index of refraction:
- for the air : n 1 = 1.00,
- for the water: n 2 = 1.33
Snell`s Law of Refraction :
n 1 · sin α 1 = n 2 · sin α 2
sin α 2 = n 1 · sin α 1 / n 2 =
= 1.00 · sin 40° / 1.33 = 0.64278 / 1.33 = 0.4833
α 2 = sin ^(-1) 0.4833
α 2 = 28.9 °
Answer: The angle relative to the water`s surface of the rays when beneath the surface is 28.9°.

4 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

Explanation:

\omega = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

Energy used in one acceleration

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J

Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
2 years ago
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