I don't understand the physics work power and energy unit

Answers

Answer 1

Answer:
Work and energy are the same thing. Their unit is the Joule.
One joule means the energy it takes to push with a force of 1 newton
through a distance of 1 meter.

The unit of power is the Watt. That's the rate of work or energy.
One watt means using 1 joule of energy (for example, push with a force of
1 newton through a distance of 1 meter), and doing that in exactly 1 second.

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During a race, a runner runs at a speed of 6 m/s. 2 seconds later, she is running at a speed of 10 m/s. What is the runner’s acceleration? Show your work.

Answers

v o = 6 m/s,
t = 2 s
v = 10 m/s
v = v o + a t
a t = v - v o
a = ( v - v o ) / t
a = ( 10 m/s - 6 m/s ) / 2 s = 4 m/s / 2 s = 2 m/s²
Answer:
The runner`s acceleration is 2 m/s².

Determine the resistance of 100m of copper cable whose cross-sectional area is 1.5mm2

Answers

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A cylindrical water tank has a radius of 3 yards in the 7.5 yards high what is the volume of the tank

Answers

To calculate the volume of a cylindrical water tank, you can use the formula for the volume of a cylinder:

Volume = π * r^2 * h

Where:

- π (pi) is approximately 3.14159

- r is the radius of the cylinder

- h is the height of the cylinder

In this case, the radius (r) is 3 yards, and the height (h) is 7.5 yards. Plug these values into the formula:

Volume = π * (3 yards)^2 * 7.5 yards

Volume ≈ 3.14159 * (9 square yards) * 7.5 yards

Now, calculate the volume:

Volume ≈ 3.14159 * 67.5 cubic yards

Volume ≈ 212.05325 cubic yards

So, the volume of the cylindrical water tank is approximately 212.05 cubic yards.

Sally travels by car from one city to another. She drives for 30.0 min at 80.0 km/hr, 12.0 min at 105.0 km/hr, and 45.0 min at 40.0 km/hr, and she spends 15.0 min eating lunch and buying gas. A. Determine the average speed for the trip.
B. Determine the total distance traveled.

Answers

Answer:

A) The average speed for the trip is $v_(average) =53,53(km)/(h)$ .

B) The total distance traveled is $d_(total)=91\ km$ .

Explanation:

Knowing that 1 hour has 60 minutes, we have that for each part of the trip:

$v_(1)=80(km)/(h)\ t_(1)=0,5\ h$

$v_(2)=105(km)/(h)\ t_(2)=0,2\ h$

$v_(3)=40(km)/(h)\ t_(3)=0,75\ h$

$v_(4)=0(km)/(h)\ t_(4)=0,25\ h$

Total distance traveled

The distance traveled is the speed multiplied by a period of time. The total distance traveled is calculated adding up the speeds multiplied by the given periods of time.

$d_(total)=d_(1)+d_(2)+d_(3)$

$d_(total)=v_(1).t_(1)+v_(2).t_(2)+v_(3).t_(3)$

$d_(total)=80(km)/(h).0,5\ h+105(km)/(h).0,2\ h+40(km)/(h).0,75\ h$

$d_(total)=40\ km+21\ km+30\ km$

$d_(total)=91\ km$

B) The total distance traveled is $d_(total)=91\ km$ .

Average speed

The average speed could be calculated using the total distance traveled $d_(total)$ divided by the total time $t_(total)$ . We need to take into account the time spent eating lunch and buying gas.

$v_(average)=(d_(total))/(t_(total))$

$v_(average)=(v_(1).t_(1)+v_(2).t_(2)+v_(3).t_(3)+v_(4).t_(4))/(t_(1)+t_(2)+t_(3)+t_(4))$ $v_(average)=(91\ km)/(0,5\ h+0,2\ h+0,75\ h+0,25\ h)$

$v_(average)=(91\ km)/(1,70\ h)$

$v_(average)=53,53(km)/(h)$

A) The average speed for the trip is $v_(average) =53,53(km)/(h)$ .

Total distance = (30/60 x 80) + (12/60 x 105) + (45/60 x 40) = 0.5 x 80 + 0.2 x 105 + 0.75 x 40 = 40 + 21 + 30 = 91 km

Average distance = total distance / time taken = 91 / (30/60 + 12/60 + 45/60) = 91/ (0.5 + 0.2 + 0.75) = 91/1.45 = 62.76 km/hr

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