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Eduardwww [97]
3 years ago
12

An electric wall clock has a second hand 15 cm long. at the tip of his hand, what is the magnitude of the velocity?

Physics
1 answer:
Mila [183]3 years ago
7 0

The velocity of the tip of the second hand is 0.0158 m/s

Explanation:

First of all, we need to calculate the angular velocity of the second hand.

We know that the second hand completes one full circle in

T = 60 seconds

Therefore, its angular velocity is:

\omega = \frac{2\pi}{T}=\frac{2\pi}{(60)}=0.105 rad/s

Now we can calculate the velocity of a point on the tip of the hand by using the formula

v=\omega r

where

\omega=0.105 rad/s is the angular velocity

r = 15 cm = 0.15 m is the radius of the circle (the distance of the point from the centre of rotation)

Substituting,

v=(0.105)(0.15)=0.0158 m/s

Learn more about angular motion here:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

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8 0
2 years ago
A 1000 kg box is being pushed with a force of 3500 N. What acceleration is the
WARRIOR [948]
Net Force = mass x acceleration
3500=1,000a
So a= 3500/1000
a=35/10
a=3.5 m/s^2
4 0
3 years ago
A ‘thermal tap’ used in a certain apparatus consists of a silica rod which
abruzzese [7]

Correct question is;

A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).

Answer:

ΔT = 268.67K

Explanation:

We are given;

d1 = 8mm

d2 = 1mm

At standard temperature and pressure conditions, the temperature is 273K.

Thus; Initial temperature; T1 = 273K,

Using the combined gas law, we have;

P1×V1/T1 = P2×V2/T2

The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:

V1/T1 = V2/T2

Now, volume of the tube is given by the formula;V = Area × height = Ah

Thus;

V1 = (πd1²/4)h

V2 = (π(d2)²/4)h

Thus;

(πd1²/4)h/T1 = (π(d2)²/4)h/T2

π, h and 4 will cancel out to give;

d1²/T1 = (d2)²/T2

T2 = ((d2)² × T1)/d1²

T2 = (1² × T1)/8²

T2 = 273/64

T2 = 4.23K

Therefore, Change in temperature is; ΔT = T2 - T1

ΔT = 273 - 4.23

ΔT = 268.67K

Thus, the temperature decreased to 268.67K

6 0
3 years ago
A machine does 500 j of work in 20 sec. What is the power of this machine?
NNADVOKAT [17]

Explanation:

P=W/t

P=500/20

P=25 W

6 0
3 years ago
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