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qwelly [4]
2 years ago
11

As Aubrey watches this merry-go-round for a total of 2 minutes, she notices the black horse pass by 15 times. What is the period

of the black horse?
Physics
2 answers:
Hunter-Best [27]2 years ago
8 0
Periodic time is the time taken for one complete oscillation by a body in circular motion. In this case the merry-go round takes 2 minutes to cover 15 complete oscillations. 2 Minutes = 120 seconds
Hence, 15 oscillations takes 120 secs
         thus 1 oscillation takes 120/15 = 8 seconds
therefore the period of the merry-go-round = 8 seconds
Grace [21]2 years ago
4 0

Answer:

8 seconds

Explanation:

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How far must 5N force pull a 50g toy car if 30J of energy are transferred?​
Alborosie

Answer: 6 m

Explanation:

30 = 5 * d

d = 30/5

d = 6 m

7 0
1 year ago
Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.4
Gre4nikov [31]

Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t  (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= (m+ \frac{1}{2} \lambda)

replacing ;

Δx = 2t   ; we have:

2t = (m+ \frac{1}{2} \lambda)

Given that thickness t = 700 nm

Then

2× 700 = (m+ \frac{1}{2} \lambda)     --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = (m+ \frac{1}{2} \lambda)     ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus;  the wavelength λ of the light when it is traveling in air = 560 nm

b)  

For the smallest thickness t_{min} ; \ \ \ m =0

∴ 2t_{min} =\frac{\lambda}{2}

t_{min} =\frac{\lambda}{4}

t_{min} =\frac{560}{4}

t_{min} =140 \ \  nm

Thus, the smallest thickness t of the air film = 140 nm

7 0
2 years ago
Read 2 more answers
A rigid, 2.50 L bottle contains 0.458 mol He. The pressure of the gas inside the bottle is 1.83 atm. If 0.713 mol Ar is added to
stellarik [79]
<span>These are inert gases, so we can assume they don't react with one another. Because the two gases are also subject to all the same conditions, we can pretend there's only "one" gas, of which we have 0.458+0.713=1.171 moles total. Now we can use PV=nRT to solve for what we want.

The initial temperature and the change in temperature. You can find the initial temperature easily using PV=nRT and the information provided in the question (before Ar is added) and solving for T.

You can use PV=nRT again after Ar is added to solve for T, which will give you the final temperature. The difference between the initial and final temperatures is the change. When you're solving just be careful with the units!
 
SIDE NOTE: If you want to solve for change in temperature right away, you can do it in one step. Rearrange both PV=nRT equations to solve for T, then subtract the first (initial, i) from the second (final, f):

PiVi=niRTi --> Ti=(PiVi)/(niR)
 
PfVf=nfRTf --> Tf=(PfVf)/(nfR)

ΔT=Tf-Ti=(PfVf)/(nfR)-(PiVi)/(niR)=(V/R)(Pf/nf-Pi/ni)

In that last step I just made it easier by factoring out the V/R since V and R are the same for the initial and final conditions.</span>
8 0
2 years ago
Read 2 more answers
A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin
Dovator [93]

Answer:

fcosθ + Fbcosθ  =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ  =Wtanθ

7 0
2 years ago
NEED ANSWER PLEASE!!!!
olganol [36]

Answer:A, Concave

Explanation:

3 0
2 years ago
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