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aksik [14]
3 years ago
11

How would you define physics class in your own words ?

Physics
1 answer:
Andrei [34K]3 years ago
3 0

Answer:

honestly i dont like physics class but for you im gonna write somethin' good but for me tho its B O R I N G

Explanation:

<em>Physics is the branch of science that deals with the structure of matter and how the fundamental constituents of the universe interact. It studies objects ranging from the very small using quantum mechanics to the entire universe using general relativity.</em>

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The maximum speed of a mass m on an oscillating spring is vmax . what is the speed of the mass at the instant when the kinetic a
umka21 [38]
Let
A =  the amplitude of vibration
k =  the spring constant
m =  the mass of the object

The displacement at time, t, is of the form
x(t) = A cos(ωt)
where
ω =  the circular frequency.

The velocity is
v(t) = -ωA sin(ωt)

The maximum velocity occurs when the sin function is either 1 or -1.
Therefore
v_{max} = \omega A
Therefore
v(t) = -V_{max} sin(\omega t)

The KE (kinetic energy) is given by
KE = \frac{m}{2}v^{2} = \frac{m}{2} V_{max}^{2} sin^{2} (\omega t)

The PE (potential energy) is given by
PE = \frac{k}{2} x^{2} = \frac{k}{2} A^{2} cos^{2} (\omega t)

When the KE and PE are equal, then
v^{2} = \frac{k}{m} A^{2} cos^{2} (\omega t)

For the oscillating spring,
\omega ^{2} =  \frac{k}{m} \\ V_{max} = \omega A =  \sqrt{ \frac{k}{m} } A
Therefore
v^{2} =  \frac{k}{m}  \frac{m}{k} V_{max}^{2} cos^{2} ( \sqrt{ \frac{k}{m} t} ) \\ v = V_{max}  \,cos( \sqrt{ \frac{k}{m} t} )

Answer: v(t) = V_{max} cos( \sqrt{ \frac{k}{m} t} )

3 0
4 years ago
An 80 kg skydiver is falling at terminal velocity. What is the value of air resistance acting on his body? Consider, what are th
Mrac [35]

Answer:

800 N

Explanation:

During terminal velocity, air resistance=weight

5 0
3 years ago
According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this
Andrew [12]

Answer:

Part a)

\rho = 0.55 g/cm^3

Part b)

\rho_L = 1.49 g/cm^3

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

Explanation:

Part a)

As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force

So here we know

buoyancy force is given as

F_b = \rho_w V_{sub} g

F_b = (1 g/cm^3) (30 - 13.5) Ag

F_b = 16.5 Ag

Now we know that the weight of the cylinder is given as

W = \rho (30 cm)A g

now we have

\rho (30 cm) A g = 16.5 A g

\rho = 0.55 g/cm^3

Part b)

When the same cylinder is floating in other liquid then we will have

F_b = \rho_L (30 - 18.9 )A g

so we have

\rho_L (11.1) Ag = 0.55(30) Ag

\rho_L = 1.49 g/cm^3

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

3 0
3 years ago
A 1650 kg car accelerates at a rate of 4.0 m/s^2. How much force is the car's engine producing?
Westkost [7]
F = M A

Force = (mass) x (acceleration) 

                   = (1,650 kg) (4 m/s²) = 6,600 kg-m/s² = <em>6,600 Newtons</em>
7 0
4 years ago
Read 2 more answers
A 5592 N piano is to be pushed up a(n) 3.79 m frictionless plank that makes an angle of 30.1 ◦ with the horizontal. Calculate th
otez555 [7]

Answer:

10628.87 J

Explanation:

We are given that

Force applied =F=5592 N

\theta=30.1^{\circ}

Displacement=D=3.79 m

We have to find the work done in sliding the piano up the plank at a slow constant rate.

Work done=F\times displacement

The perpendicular component of force=FSin\theta=5592sin(30.1)=2804.45N

Work done =Fsin\theta\times D=2804.45\times 3.79=10628.87 J

Hence, the work done in sliding the piano up the plank at a slow constant rate=10628.87 J

8 0
3 years ago
Read 2 more answers
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