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1 year ago

What is the period and frequency of the second hand on a clock?

Physics

1 answer:

Tanzania [10]1 year ago

6 0

Answer:

Explanation:For second hand, period = 1 min = 3660 s; frequency = 0.0166 Hz}\\\mbox{For minute hand, period = 1 hr = 3600 s; frequency = 2.77}\times10^{-4} Hz\\\mbox{For hour hand, period = 12 hrs = 4300 s; frequency = 2.3}\times10^{-5} Hz.

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In a bi-prism experiment the eye-piece was placed at a distance 1.5m from the source. The distance between the virtual sources w

Vladimir [108]

Answer:

λ = 1.4 × 10^(-7) m

Explanation:

We are given;

distance of eye piece from the source;D = 1.5 m

distance between the virtual sources;d = 7.5 × 10^(-4) m

To find the wavelength, we will use the formula for fringe width;

X = λD/d

Where X is fringe width, λ is wavelength, while d and D remain as before.

Now, fringe width = eye-piece distance moved transversely/number of fringes

Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m

Thus,

Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m

Thus;

1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))

λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5

λ = 1.4 × 10^(-7) m

6 0

2 years ago

. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110

7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x $10^{5}$ J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, $v_{0}$ = 110 km/h = 110 x 1000/3600 = 30.6 m/s $\frac{30.6^{2} }{2x9.8}$

initial height, $h_{0}$ = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = $\frac{1}{2} mv_{0} ^{2}$

gh = $\frac{1}{2} v_{0} ^{2}$

h = $\frac{v_{0} ^{2} }{2g}$

= 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

= mg (h - $h_{0}$ )

= 750 x 9.8 x (47.6 - 22)

= 188160 Joule

= 1.88 x $10^{5}$ J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

= 1.88 x $10^{5}$ /(22/sin 2.5°)

= 373 N

8 0

2 years ago

The capacitor is now reconnected to the battery, and the plate separation is restored to d. A dielectric plate is slowly moved i

Paha777 [63]

Answer:

U2 = KAε0V2 / (2d)

Explanation:

The dielectric constant K just replaces the "3″ from Part B.

8 0

2 years ago

what should be the angle between two vectors of same angle so that their resultant is equal to either of them?

asambeis [7]

So, the angle between two vectors having equal magnitude is equal to 120º.

Explanation:

<h2>:) Correct me if I'm wrong...</h2>

3 0

1 year ago

A small sphere with a mass of 441 g is moving upward along the vertical +y-axis when it encounters an electric field of 5.00 N/C

sweet-ann [11.9K]

In your question where the ask is to calculate the charge that the small sphere carries which is the mass of it is 441g moving at an acceleration of 13m/s^2 nad having and electric field of 5N/C. So the formula in getting the charge is mutliply the mass and the quotients of Acceleration and the Electric Field so the answer is 1,146.6

3 0

2 years ago