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notsponge [240]
3 years ago
14

Bonus: What is the velocity of an 8 kg lead shot-put if it has 484 J of energy? Type here to search e​

Physics
1 answer:
Dima020 [189]3 years ago
7 0

Answer:

11m/s

Explanation:

Given parameters:

Mass of the lead shot = 8kg

Energy of the shot  = 484J

Unknown:

Velocity of the shot  = ?

Solution:

To solve this problem, we apply the kinetic energy formula;

        K.E  = \frac{1}{2} m v²  

m is the mass

v is the unknown

 Now, insert the parameters and solve;

        484  = \frac{1}{2} x 8 x v²

        484  = 4v²

           v² = 121

            v  = 11m/s

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A uniform, 0.0300-kg rod of length 0.400 m rotates in a horizontal plane about a fixed axis through its center and perpendicular
AURORKA [14]

Answer:

A) ω₂ = 28 rev/min

B) ω_3 = 28 rev/min

Explanation:

A) Initial moment of inertia (I₁) is the rod's own ((1/12)ML²) plus that of the two rings (two lots of mr² if we treat each ring as a point mass).

Thus;

I₁ = ((1/12)ML²) + 2(mr²)

Where;

M is mass of rod = 0.03 kg

L is length of rod = 0.4m

m is mass of small rings = 0.02 kg

r is radius of small rings = 0.05 m

Thus;

I₁ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.05²)

I₁ = 0.0021 kg.m²

Now, when the rings slide and reach the ends of the rod i.e at r = L/2 = 0.4/2 = 0.2m from axis, the new Moment of inertia is;

I₂ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.2²)

I₂ = 0.0036 kg.m²

From conservation of angular momentum, we know that:

I₁ω₁ = I₂ω₂

We are given ω₁ = 48 rev/min.

Thus; plugging in the relevant values;

0.0021 x 48 = 0.0036ω₂

0.1008 = 0.0036ω₂

ω₂ = 0.1008/0.0036

ω₂ = 28 rev/min

b) For this, we have the same scenario as the case above where the ring just reaches the ends of the rods.

Thus,

I₂ω₂ = I_3•ω_3

So,0.0021 x 48 = 0.0036ω_3

ω_3 = 28 rev/min

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3 years ago
A block with mass M = 3 kg is moving on a flat surface with constant speed v1 =
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Answer:

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Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a p slope at constant speed, as sho
kondor19780726 [428]

The work done by friction to move the sled is  - 1,323 J.

<h3>What is Coefficient of friction?</h3>
  • The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them.
  • Typically, it is represented by the Greek letter µ. In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
  • The coefficient of friction has no dimensions because both F and N are measured in units of force (such as newtons or pounds). For both static and kinetic friction, the coefficient of friction has a range of values.
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Solution:

Given that

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We use the formula:

friction work = -µmgdcos∅

friction work = -0.100 × 90 kg × 9.8 m/s² × 30 m × cos 60°

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Know more about Coefficient of friction numerical brainly.com/question/19308401

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Answer:

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Explanation:

We have,

Heat required in aluminium to change the temperature from 68°C to 110°C. It is required to find the mass of aluminium.

Concept used : Specific heat capacity

Solution,

The heat required to raise the temperature is given by :

Q=mc\Delta T

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So, the mass of aluminium is 105.58 grams.

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