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Kipish [7]
3 years ago
13

What is the length of the shadow cast on the vertical screen by your 10.0 cm hand if it is held at an angle of θ=30.0∘ above hor

izontal? Express your answer in centimeters to three significant figures. View Available Hint(s)
Physics
1 answer:
Alchen [17]3 years ago
5 0

Answer:

The  length is  D  =  5 \ cm

Explanation:

From the question we are told  that

     The  length of the  hand is  l  =  10.0 \ cm

      The  angle at the hand is  held is  \theta  =  30 ^o

Generally resolving the length the length of the hand to it vertical component we obtain that the length of the shadow on the vertical wall is mathematically evaluated as

             D  =  l * sin(\theta )

substituting values

             D  =  10 * sin (30)

             D  =  5 \ cm

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4 0
3 years ago
What does it mean to say that a wave is a cyclic, or periodic, disturbance
Ivenika [448]

Wave is a disturbance or energy that propagate through medium from one point to other point

So basically it is a flowing energy that flows into the medium and hence medium particles start oscillating about their mean position to and fro.

This motion of medium particles or this to and fro motion is about their mean position and this will always be cyclic or periodic motion

This means the disturbance or energy continuously flow through the medium such that it will change the position of medium particle and this will be cyclic in order

For an example

y = A sin(\omega t + kx + \phi)

so here above equation of wave is a travelling wave in which displacement of medium particle from its mean position is given by "y"

Now we can see that this disturbance depends upon the sine function and it will repeat its same position after every 2 pi time interval as it is cyclic function for this value

Due to this phenomenon of repeatation of its same position we can say that it is disturbance of wave is cyclic.

3 0
3 years ago
A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately
Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car V_{c} = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

V_{f}  = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

V_{f} =  V_{i} + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁  = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁  will be

d_{c} = V_{c} × t₁

we substitute

d_{c} = 22.352 × 7

d_{c}  = 156.46 m

now distance between polive car and speeding car

Δd =  d_{c} - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( V_{f} - V_{c} )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = V_{f} × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0
3 years ago
A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function
Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$v(t)=\sin (\omega t) * \cos ^2(\omega t)

To get the position equation we just need to integrate the above equation:

$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t

$\mathrm{u}=\cos (\omega \mathrm{t})

Then:

$d u=-\sin (\omega t) d t

\Rightarrow d t=-d u / \sin (\omega t)

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$f(t)=\frac{\cos ^3(\omega t)}{3}+C

To find the value of C, we use the fact that f(0) = 0.

$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0

C = -1 / 3

Then the position function is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

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