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navik [9.2K]
3 years ago
15

In a device like the one shown below, the cylinder is allowed to fall a distance of 300 m. As a result, the temperature of the w

ater increases by 2.7°C. What would have been the increase in temperature if the cylinder were only allowed to fall 100 m?
Physics
1 answer:
Delvig [45]3 years ago
8 0

Answer:

Increase in the temperature of water would be 0.9 degree C

Explanation:

As we know by energy conservation

Change in the gravitational potential energy of the cylinder = increase in the thermal energy of the water

Here we know that the gravitational potential energy of the cylinder is given as

U = mgh

here we have

h = 300 m

now we can say

Mc\Delta T = (m \times 9.8 \times 300)

now if the cylinder falls from height h = 100 m

then we have

Mc\Delta T' = (m \times 9.8 \times 100)

now from above two equations

\frac{\Delta T'}{\Delta T} = \frac{100}{300}

\Delta T' = 2.7 \times \frac{1}{3} = 0.9 Degree C

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A 1500 kg weather rocket accelerates upward at 10.0 m/s . It explodes 2.00 s after liftoff and breaks into two fragments, one tw
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20 m/s is the speed of the heavier fragment just after the explosion.

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

F = ma                   where, F = Force (Newton)

                                          m= mass

                                          a = acceleration

Given:

mass of rocket = M = 1500 kg

acceleration of rocket = a = 10 m/s²

elapsed time = t = 2.00 s

mass of lighter fragment = m₁ = m = 500 kg

mass of heavier fragment = m₂ = 2m = 1000 kg

maximum height of lighter fragment = h = 530 m

Let's calculate the final speed of the rocket just before the explosion:

v = u + at

v = 0 + 10(2)

v = 20 m/s

Then, we will calculate the height of the rocket just before the explosion:

h' = ut + \frac{1}{2}at^{2}

h' = 0 + \frac{1}{2} (10)(2.00)^{2}

h' =20m

The initial speed of lighter fragment just after the explosion:

v^{2}₁ = u^{2}₁ - 2gΔh

v^{2}₁ = u^{2}₁ - 2g (h - h')

0^{2} = u^{2}₁- 2(9.8) (530-20)

u^{2}₁=9996

u₁ =\sqrt{9996}  m/s

Using Conservation of Momentum Law :

M v=m₁ u₁ +  m₂u₂

1500 (20)= 500(\sqrt{9996} ) + 1000u₂

u₂ ≅ - 20 m/s

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1 year ago
True or false? Scientific laws can be expressed through a law that relates several variables.
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Answer:

True

Explanation:

Scientific laws are often written as expressions that contains variables and are laws that are binding themselves.

In science, laws are natural phenomenon that draws from careful observations that holds through following a series of detailed study. Within the range of assumed parameters, a law will always hold true.

Most laws in science are denoted using mathematical variables which helps to interpret them.

The variables shows the relationship between the different parts of the law.

For example, Newton's law of universal gravitation is expressed mathematically as shown below;

                          F  =  \frac{G m_{1} m_{2}  }{r^{2} }

where G, m and r are all variables.

  G is the universal gravitation constant

   m is mass

   r is the distance between them.

   F is the gravitational force.

Most scientific laws are often expressed in this format.

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2 years ago
in case one, a car speeds up from zero m/s to 15 m/s. in case two, the same car speeds up from 15 m/s to 30 m/s. the mass of the
GrogVix [38]

Three time  the energy was needed than in case one.

What is energy?

In physics, energy is the quantifiable property that is transmitted to a body or a physical system and is visible in the form of heat and light. The law of conservation of energy states that energy can be converted in form but cannot be created or destroyed.

Case 1:

Vo = 0 m/s

V1 = 15 m/s

Case 2:

Vo = 15 m/s

V1 = 30 m/s

Change in KE = 1/2 × m × (V1^2 - Vo^2)

KE1 = 1/2 × 1000 × (15^2 - 0)

= 112.5 kJ

KE2 = 1/2 × 1000 × (30^2 - 15^2)

= 1/2 × 1000 × 675

= 337.5kJ

Case 1 has an increase of 112.5 kJ energy while Case 2 has an increase of 337.5kJ.

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A relief plane flying horizontally at 65.0 m/s is delivering a
SVETLANKA909090 [29]

The height of the plane delivering a food package to people stranded on an island is 722.2 m.

<h3>Time of motion of the food package</h3>

The time taken for the food package to hit the ground level is calculated as follows;

X = vt

where;

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  • v is horizontal distance
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t = X/v

t = 789/65

t = 12.14 seconds

<h3>Height of the plane</h3>

h = ¹/₂gt²

where;

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  • t is time of motion of food package
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h = (0.5)(9.8)(12.14²)

h = 722.2 m

Thus, the height of the plane delivering a food package to people stranded on an island is 722.2 m.

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1 year ago
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