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1 year ago

A school bus is moving 26.8 m/s on

Physics

1 answer:

Elodia [21]1 year ago

3 0

The time taken for the bus to travel 95.1 m is 2.8 s

<h3>How to determine the final velocity</h3>

Initial velocity (u) = 26.8 m/s
Acceleration (a) = 4.73 m/s²
distance (s) = 95.1 m
Final velocity (v) = ?

v² = u² + 2as

v² = 26.8² + (2 × 4.73 × 95.1)

v² = 1617.886

Take the square root of both sides

v² = √1617.886

v = 40.2 m/s

<h3>How to determine the time</h3>

Initial velocity (u) = 26.8 m/s
Acceleration (a) = 4.73 m/s²
Final velocity (v) = 40.2 m/s
Time (t) = ?

v = u + at

40.2 = 26.8 + (4.73 × t)

Collect like terms

4.73 × t = 40.2 - 26.8

4.73 × t = 13.4

Divide both side by 4.73

t = 13.4 / 4.73

t = 2.8 s

Learn more about velocity:

brainly.com/question/3411682

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A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface: