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ELEN [110]
4 years ago
15

A race car starts from rest on a circular track of radius 478 m. The car’s speed increases at the constant rate of 0.747 m/s 2 .

At the point where the magnitudes of the centripetal and tangential accelerations are equal, find the speed of the race car
Physics
1 answer:
Keith_Richards [23]4 years ago
3 0

Answer:

18.896ms-1

Explanation:

The tangential acceleration of the car= 0.747ms-2

The centripetal acceleration is given by v^2/r

0.747= v^2/r

V= √0.747*478= 18.896ms-1

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At 2 P.M., ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast i
labwork [276]

Answer:

The distance between the ships changing at 6PM is 21.29Km/h

Explanation:

Ship A is sailing east at 35Km/h and ship B is sailing West at 25Km/h

Given

dx/dt= 35

dy/dt= 25

dv/dt= ???? at t= 6PM - 2PM= 4

Therefore t=4

We know ship A travels at 150km in the x-direction and Ship A at t=4 travels at 4.35 Which is 140 also in x-direction

So, we use:

D^2 = (150 - x)^2 + y^2;

D^2 = (150 - 140)^2 + y^2

But ship B travels at t=4, at 4.25 =100 in the y-direction

so, let's use the equation:

D^2 = 10^2 + 100^2

= D= sqrt*(10 + 100)

Lets use 2DD' = 2xx' + 2yy'

Differentiating with respect to t we have:

D•d(D)/dt = -(10)•dx/dt + 100•dy/dt

=100.5 d(D)/dt = (-10)•35 + (100)•25

When t=4, we have x=(140-150) =10 and y=100

= D = sqrt*(10^2 + 100^2)

=100.5

= 100.5 dD/dt = 10.35 +100.25

= dD/dt = 21.29km/h

7 0
3 years ago
Calculate the frequency of each of the following wave lengths of electromagnetic radiation. Part A 488.0 nm (wavelength of argon
Drupady [299]

Answer:

A. f=6.1475*10^{14}Hz

B. f=5.9642*10^{14}Hz

Explanation:

The frequency has an inversely proportional relationship with the concept of wavelength, the greater the wavelength, the lower the frequency. For electromagnetic waves, the frequency is equal to the speed of light, divided by the wavelength.

f=\frac{c}{\lambda}

A.

f=\frac{3*10^8\frac{m}{s}}{488*10^{-9}m}\\\\f=6.1475*10^{14}Hz

B.

f=\frac{3*10^8\frac{m}{s}}{503*10^{-9}m}\\\\f=5.9642*10^{14}Hz

7 0
4 years ago
If the ultraviolet photon has a wavelength of 249 nm and one of the photons emitted by the fluorescent material has a wavelength
melamori03 [73]

Answer:

601 nm

Explanation:

Energy of photon having wavelength of λ nm

= \frac{1244}{\lambda}eV

Energy of 249 nm photon

=\frac{1244}{249}

=4.996 eV

Similarly energy of 425nm photon

=\frac{1244}{425}

=2.927 eV

Difference = 2.069 eV.

This energy will give rise to another photon whose wavelength will be

λ = \frac{1244}{2.069}

= 601 nm.

6 0
4 years ago
Higher-pitched sounds have shorter wavelengths.<br><br> True or false
Eva8 [605]

Answer:

The answer is True

Explanation:

I hope it helps

5 0
2 years ago
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In a standing wave, points that remain fixed are the
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The answer is nodes because nodes stay in a fixed position and i just learnt about this! 
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