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Verdich [7]
3 years ago
9

5. A body travels in a circle at a constant speed, it

Physics
2 answers:
Scilla [17]3 years ago
7 0
No acceleration as the term acceleration describes change in velocity so if its speed is same there will be no acceleration
nata0808 [166]3 years ago
5 0
A the same acceleration
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What is the massof the largest ruby?
alexandr402 [8]
I think the answer is 2283g
4 0
3 years ago
A positive test charge of 8.5 × 10 negative 7 Columbus experiences a force of 4.1 × 10 negative 1 N calculate the electric field
Sophie [7]

Explanation:

direction of electric field is same as that of force experienced by the test charge

8 0
2 years ago
A double-slit experiment uses light of wavelength 650 nm with a slit separation of 0.100 mm and a screen placed 4.0 m away. a) W
dezoksy [38]

Answer:

Explanation:

a ) Slit separation d = .1 x 10⁻³ m

Screen distance D = 4 m

wave length of light  λ = 650 x 10⁻⁹ m

Width of central fringe = λ D / d

= \frac{650\times10^{-9}\times4}{.1\times10^{-3}}

= 26 mm

b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm

c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to

λ  / d

= \frac{650\times10^{-9}}{.1\times10^{-3}}

= 6.5 x 10⁻³ radian.

8 0
3 years ago
Ow much charge flows from a 12.0 v battery when it is connected across a completely discharged 18.0 μf capacitor
Trava [24]
The equation Q=CV (Charge = product of Capacitance and potential difference) tells us that the maximum charge that can be stored on a capacitor is equal to the product of it's capacitance and the potential difference across it. In this case the potential difference across the capacitor will be 12.0V (assuming circuit resistance is negligable) and it has a capacitance of 18.0μf or  18.0x10^-6f, therefore charge equals (18.0x10^-6)x12=2.16x10^-4C (Coulombs).
5 0
2 years ago
An air bubble has a volume of 2.0 cm3 when it is released by a submarine 100 m below the surface of a freshwater lake. What is t
UkoKoshka [18]

Answer:

21.35 cm^3

Explanation:

let the volume at the surface of fresh water is V.

The volume at a depth of 100 m is V' = 2 cm^3

temperature remains constant.

density of water, d = 1000 kg/m^3

Pressure at the surface of fresh water is atmospheric pressure,

P = Po = 1.013 x 10^5 N/m^2

The pressure at depth 100 m is P' = Po + hdg

P' = 1.013 \times 10^{5}+ 100 \times 1000 \times 9.8

P' = 10.813 x 10^5 N/m^2

Use the Boyle's law

P V = P' V'

1.013 \times 10^{5}\times V = 10.813 \times 10^{5}\times 2

V = 21.35 cm^3

Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.

5 0
3 years ago
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