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krek1111 [17]
2 years ago
10

A 20-kg boy slides down a smooth, snow-covered hill on a plastic disk. The hill is at a 10° angle to the horizontal, and the slo

pe is 50 m long. Part A If the boy starts from rest, what is his speed at the bottom of the hill?​
Physics
1 answer:
zvonat [6]2 years ago
7 0

Answer:

13 m/s

Explanation:

I assume we are ignoring friction.

The boy's PE will all be converted to KE at the bottom of the hill.

to find PE = mgh   we need to know h

   h = 50 sin 10 = 8.68 meters

     then:    PE = 20 * 9.81 * 8.68 =<u> 1703.49</u> j

KE = 1/2 m v^2 = <u>1703 .49</u>

            v = 13 m/s

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For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
3 years ago
A sound has a sound level of 30 dB. Its intensity is what multiple of the standard reference level for intensities?
arsen [322]
<h2>Answer:</h2>

1000th multiple of the standard reference level for intensities.

<h2>Explanation:</h2>

The sound intensity level (β), measured in decibels, of a sound with an intensity of I is defined as follows;

β = 10 log (I / I₀)       --------------------(i)

Where;

I₀ = reference intensity

Given from the question;

β = sound level = 30dB

Substitute this value into equation (i) as follows;

30 = 10 log (I / I₀)

Divide both sides by 3;

3 = log (I / I₀)

Take antilog of both sides;

10^(3) = (I / I₀)

1000 = I / I₀

Solve for I;

I = 1000I₀

Therefore the intensity of the sound is 1000 times the standard reference level for intensities (I₀)

7 0
3 years ago
An ore sample weighs 17.50 N in air. When the sample
zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T=W-F_b

F_b=buoyancy force

T=Tension force

W=Weight

By using the formula

11.2=17.5-F_b

F_b=17.5-11.2=6.3 N

F_b=V_{object}\times \rho_{water}\cdot g

Where

V_{object}=Volume of object

\rho_{water}=1000 kgm^{-3}=Density of water

g=9.8 ms^{-2}=Acceleration due to gravity

Substitute the values then we get

6.3=9.8\times 1000\times V_{object}

V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3

Volume of sample=6.43\times 10^{-4} m^3

Density of sample,\rho_{object}=\frac{Mass}{volume_{object}}

Where mass of ore sample=1.79 kg

Substitute the values then, we get

\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3

Density of the sample=2.78\times 10^{3} kgm^{-3}

7 0
3 years ago
When a board with a box on it is slowly tilted to a larger and larger angle, common experience shows that the box will at some p
Shtirlitz [24]

Answer:

the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic frictionExplanation:

This exercise uses Newton's second law with the condition that the acceleration is zero, by the time the body begins to slide. At this point the balance of forces is

    fr- w || = 0

The expression for friction force is that it is proportional to the coefficient of friction by normal.

    fr = μ N

When the system is immobile, the coefficient of friction is called static coefficient and has a value, this is due to the union between  the surface, when the movement begins some joints are broken giving rise to coefficient of kinetic friction less than static.  

In consequence a lower friction force, which is why the system comes out of balance and begins to accelerate.

      μ kinetic <μ static

  In all this movement the normal with changed that the angle of the table remains fixed.

Consequently, the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic friction

3 0
3 years ago
A bullet is fired into the air at an angle of 30° to the horizontal. At the same time and from the same height,
enyata [817]

Answer:

The bullet that is fired will spend longer  in the air, hitting the ground after the  dropped bullet.

Explanation:

Using the equation: x =  x 0  +   v t

If we neglect the effects of air resistance, the horizontal motion is a constant velocity.

The horizontal displacement = (velocity X cosθ)

So, the fired bullet has to travel horizontally before falling which takes a  longer time compared to a bullet dropped where it is, height = 1/2 gt^2

gravity, g = 9.8 m/s2.

3 0
2 years ago
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