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Marianna [84]
4 years ago
13

What is the angular speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch?

answer in radians per second?
Physics
1 answer:
Agata [3.3K]4 years ago
8 0
A)We know the formula of the angular speed is                 ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is          ω= 2π / (60s)             = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is             T = 1 hr                = 3600sThen the angular speed of the minute hand is         ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is          T = 12hrs             = (12)(3600)sThen the angular speed of the hour hand is         ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s
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How to calculate what?
6 0
4 years ago
You are given a copper bar of dimensions 3 cm × 5 cm × 8 cm and asked to attach leads to it in order to make a resistor. If you
Nataliya [291]

Answer:

Option B is correct.

The leads should be attached to the opposite faces that measure 3 cm × 5 cm to obtain the maximum possible resistance.

Explanation:

The resistance of a material is given by

R = (ρL)/A

where ρ = resistivity of the material

A = Cross sectional Area through which current would flow

L = length of the material.

From the relation, it is evident that the resistance of a material is directly proportional to its length and inversely proportional to its cross sectional Area.

As the length of material increases, the resistance of the material also increases, and a decreasing length translates to a decreasing resistance too. (Direct Proportionality)

And as the cross sectional Area of the material increases, the resistance of the material decreases. Decrease in cross sectional Area of the material translates to an increase in resistance. (Inverse Proportionality)

To maximize resistance of a material, it would make sense to maximize the length and minimize the cross sectional Area of that material. (Since resistivity is assumed to be constant)

And with a dimension of (3 × 5 × 8), the longest length is 8 cm and the smallest cross sectional Area is (3 × 5).

So, the leads should be atrached to the face with area (3 cm × 5 cm), which is the smallest cross sectional Area and gives the largest length between the faces (8 cm).

This subsequently maximizes the resistance.

Hope this Helps!!!

4 0
3 years ago
Read 2 more answers
What is the magnitude of velocity for a 2,000 kg car possessing 3,000 kg(*)m/s of momentum?
Trava [24]

The magnitude of velocity for this car is equal to 1.5 m/s.

<u>Given the following data:</u>

  • Momentum of car = 3,000 kgm/s.
  • Mass of car = 2,000 kg.

To calculate the magnitude of velocity for this car:

<h3>What is momentum?</h3>

In Science, momentum simply means a multiplication of the mass of an object and its velocity.

Mathematically, momentum is giving by the formula;

Momentum = mass \times velocity

Making velocity the subject of formula, we have:

Velocity=\frac{Momentum}{Mass}

Substituting the given parameters into the formula, we have;

Velocity=\frac{3000}{2000}

Velocity = 1.5 m/s.

Read more on momentum here: brainly.com/question/15517471

5 0
3 years ago
When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and a
ryzh [129]

Complete question:

The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?

Answer:

The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

Explanation:

Given;

combined mass of the shotgun and arm–shoulder, m₁ = 15 kg

mass of the projectile, m₂ = 0.04 kg

speed of the projectile, u₂ = 380 m/s

let the recoil velocity of the shotgun and arm–shoulder combination = u₁

Apply the principle of conservation of linear momentum;

m₁u₁  +  m₂u₂ = 0

m₁u₁ = - m₂u₂

u_1 = -\frac{m_2u_2}{m_1} \\\\u_1 = - \frac{0.04\times 380}{15} \\\\u_1 =-1.013 \ m/s\\\\u_1 = 1.013 \ m/s \ \ \ in \ opposite \ direction

Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

3 0
4 years ago
A man starts his car from rest and accelerates at 1 m/s2 for 2 s. He then continues at a constant velocity for 10 s until
Bumek [7]

Answer: I showed you all calculation . You did not attach any graph to question .

Explanation:

Lets first find Velocity

Vr=o m/s

Ve=?

a=1m/s²

t=2s

----------

a=(Vr-V)/t

1m/s²=Vr-0m/s/2s

2m/s=Vr

Lets find the time neeeded to stop :

a=1m/s²

Vs=2m/s

Vf=0m/s

a=(Vf-Vs)/t

t*1m/s²=2m/s

t=2 s

4 0
3 years ago
Read 2 more answers
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