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scoray [572]
3 years ago
11

At highway speeds, a particular car is capable of an acceleration of about 1.6 m/s?. At this rate,

Physics
1 answer:
kicyunya [14]3 years ago
5 0

Given parameters:

Acceleration of the car = 1.6m/s

Initial speed  = 80km/hr

Final speed  = 110km/hr

Solution:

Time taken to achieve this speed = ?

Solution:

Acceleration is the rate of change of velocity with the time taken.

  Mathematically;

     a  = \frac{V - U}{T}

where a is the acceleration

           V is the final velocity

           U is the initial velocity

           T is the time taken

Now make the unknown time the subject of the expression;

      aT  = V - U

         T = \frac{V - U}{a}  

Convert the given acceleration to km/hr;

       1.6m/s  = 1.6 x \frac{m}{s}  x \frac{1km}{1000m} x \frac{3600s}{1hr} = 5.76km/hr

Input the parameters and solve;

       T = \frac{110 - 80}{5.76}  = 5.2hrs

The time taken is 5.2hrs

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Answer:

3.55kg

Explanation:

K. E = ½mv²

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K. E = 300J

v = 13m/s

Therefore,

K. E = ½mv²

300 = ½ × m × 13²

Multiply both sides by 2

600 = m × 169

Divide both sides by 169

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How high does Pete lift his sledge hammer if he used a force of 25N to lift the hammer while doing 50J of work?
sasho [114]

Answer:He lifts 2 meters

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A crew team rows a boat at a rate of 20 km/h in still water. In practice on a river, the team rows for 30 minutes up the river (
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In the second 30 mins, the speed should be 20 + 1.5 = 21.5 km/h

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3 years ago
How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

T_{f} =  (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

T_{f}  = 466.8 K

so change in temperature ΔT

ΔT =  466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x_{i\\}² - x_{f\\}² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q =  16 J

Therefore, the required heat energy is 16 J

5 0
2 years ago
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