Answer:
No.
Explanation:
Given that Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change ?
Considering the formula
V = wr
Where
V = linear speed
W = angular speed
r = radius of the wheel.
But W = 2πrf
Where the the 2 and pi are constant. The radius of the first wheel will be small but counter balance with the larger frequency.
While the radius of the second wheel may be large but it will be of a small frequency.
We can therefore conclude that the reading on the speedometer will not change. Because speedometer will read the linear speed V.
Answer:
Q = 1057.5 [cal]
Explanation:
In order to solve this problem, we must use the following equation of thermal energy.
where:
Q = heat energy [cal]
Cp = specific heat = 0.47 [cal/g*°C]
T_final = final temperature = 32 [°C]
T_initial = initial temperature = 27 [°C]
m = mass of the substance = 450 [g]
Now replacing:
![Q=450*0.47*(32-27)\\Q=1057.5[cal]](https://tex.z-dn.net/?f=Q%3D450%2A0.47%2A%2832-27%29%5C%5CQ%3D1057.5%5Bcal%5D)
Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
Answer:
The answer is D
Explanation:
I'm too lz to explain everything.
sorry.
The power of man performing 500 J of work in 8 seconds is 62.5 J/s.
Power can be defined as the pace at which work is completed in a given amount of time.
Horsepower is sometimes used to describe the power of motor vehicles and other machinery.
The pace at which work is done on an item is defined as its power. Power is a temporal quantity.
Which is connected to how quickly a project is completed.
The power formula is shown below.
Power = Energy / Time
Power = E / T
Because the standard metric unit for labour is the Joule and the standard metric unit for time is the second, the standard metric unit for power is a Joule / second, defined as a Watt and abbreviated W.
Here we have given Energy as 500 J and Time as 8 second.
Power = Energy / Time
Power = 500 / 8 Joule / sec
Power = 250 / 4 Joule / sec
Power = 125 / 2 Joule / sec
Power = 62.5 Joule / sec or 62.5 watt
Power came out to be 62.5 J/s when the man performed 500 Joule of work in 8 seconds.
So we can conclude that the power in the Energy transmitted per unit of time, and can be find out by dividing Energy by time. In our case the Power came out to be 62.5 Joule / Second.
Learn more about Power here:
brainly.com/question/1634438
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