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Otrada [13]
3 years ago
12

A child and sled with a combined mass of 50 kg, start from rest and slide down a frictionless hill that is 7.5 meters high. what

is the mechanical energy of the sled at the top? what is the mechanical energy of the sled at the bottom? what is the speed of the sled at the bottom of the hill
Physics
1 answer:
amid [387]3 years ago
3 0
The mechanical enegia is the sum of the kinetic energy plus the potential energy
 Kinetic energy = (1/2) * m * v ^ 2
 Potential energy = m * g * h
 Mechanical energy = (1/2) * m * v ^ 2 + m * g * h
 What is the mechanical energy of the sled at the top? 
 Mechanical energy = (1/2) * m * v ^ 2 + m * g * h
 Mechanical energy = (1/2) * (50) * (0) ^ 2 + (50) * (9.8) * (7.5) = 3675
 Mechanical energy = 3675J
 What is the mechanical energy of the sled at the bottom? 
 By conservation of energy we have that the energy in point 1 is equal to the energy in point 2
 Mechanical energy = 3675J
 What is the speed of the sled at the bottom of the hill?
 Mechanical energy = 3675J = (1/2) * m * v ^ 2
 clearing up v we have
 (1/2) * (50) * v ^ 2 = 3675
 v ^ 2 = 3675 * (2) * (1/50)
 v = root (3675 * (2) * (1/50)) = 12.12 m / s
 answer
 3675J
 3675J
 12.12 m / s
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Answer: You do not specify what is being asked for. ∆E? ∆H?

∆E = (430 - 238) J = 192 J

∆H = 430 J

Explanation:

If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.

Therefore ∆H = 430 J

If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w

The question states that 238 J of work are done AND the system expanded

(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)

Therefore, ∆E = (430 - 238) J = 192 J

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A car driving on the highway is going 72mph After 3 hours how many miles will the car have driven ?
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Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i
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Answer:

<em>a) 0.72 V</em>

<em>b) 19.2 mA</em>

<em>c) 2.304 Watts</em>

Explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns = N_{p} = 500 turns

input voltage = V_{p} = 120 V

number of secondary turns = N_{s} = 3 turns

output voltage = V_{s} = ?

using the equation for a transformer

\frac{V_{s} }{V_{p} }  = \frac{N_{s} }{N_{p} }

substituting values, we have

\frac{V_{s} }{120 }  = \frac{3 }{500} }

500V_{p}  = 120*3\\500V_{p} = 360

V_{p} = 360/500 =<em> 0.72 V</em>

<em></em>

b) by law of energy conservation,

I_{P}V_{p} = I_{s}V_{s}

where

I_{p} = input current = ?

I_{s} = output voltage = 3.2 A

V_{s} = output voltage = 0.72 V

V_{p} = input voltage = 120 V

substituting values, we have

120I_{p} = 3.2 x 0.72

120I_{p} = 2.304

I_{p}  = 2.304/120 = 0.0192 A

= <em>19.2 mA</em>

<em></em>

c) power input = I_{p} V_{p}

==> 0.0192 x 120 = <em>2.304 Watts</em>

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