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joja [24]
3 years ago
7

What is the average temperature of the city, 89.6 Fahrenheit, on the Celsius scale?

Physics
1 answer:
DaniilM [7]3 years ago
8 0

Answer:

thrity-two degrees Celsius

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A thin film of oil of thickness t is floating on water. The oil has index of refraction no = 1.4. There is air above the oil. Wh
kkurt [141]

Answer:

t = 120.5 nm

Explanation:

given,    

refractive index of the oil = 1.4

wavelength of the red light = 675 nm

minimum thickness of film = ?

formula used for the constructive interference

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

where n is the refractive index of oil

t is thickness of film

for minimum thickness

m = 0

2 \times 1.4 \times t = (0+\dfrac{1}{2})\times 675

t = \dfrac{0.5\times 675}{2\times 1.4}

        t = 120.5 nm

hence, the thickness of the oil is t = 120.5 nm

7 0
3 years ago
What is meant by radioactivity?
vladimir1956 [14]
What that means is the atom is so radioactive that the nucleus is unstable.

4 0
2 years ago
Help please:Calculate the mechanical energy of a bird flying at a speed of 10 m / s at an altitude of 15 m. If its mass is 150 g
Sonja [21]

Mass of the bird(m) = 150 g = 0.15 kg

Speed (v) = 10 m/s

Kinetic Energy = \frac{1}{2}m v^{2} = \frac{1}{2} 0.15 (10)^{2} = 7.5 J

Altitude (h) = 15 m

Gravitational Potential Energy = (0.15)(9.81)(15) = 22.0725 J

Mechanical Energy = Kinetic Energy + Potential Energy = 7.5 + 22.0725

= 29.5725 J

4 0
2 years ago
Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire
Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

F = ma \rightarrow a = \frac{F}{m}

Replacing,

a =\frac{42N}{83000kg}

a =5.06*10^{-4}m/s^2

The total speed change

\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity } we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}

We have that,

t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }

t = 1403.16s

Therefore the Rocket should be fired around to 1403.16s

7 0
3 years ago
HELP ME please
allochka39001 [22]

Answer:

Ruko zara kuch Time dedo na please

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