True or false. Students with a Learners License may not receive a motorcycle endorsement.

skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

v (ave) = ∆x / ∆t

and under constant acceleration,

v (ave) = (v (final) + v (initial)) / 2

According to the plot, with ∆t = 4.0 s, we have v (initial) = 0 and v (final) = 10.0 m/s, so

∆x / (4.0 s) = (10.0 m/s) / 2

∆x = ((4.0 s) • (10.0 m/s)) / 2

∆x = 20.00 m

5 0

2 years ago

In what region of the electromagnetic spectrum is a photon found that possesses twice as much energy as one in the blue region (

iren2701 [21]

The answer is: ultraviolet The energy (E) of a photon is directly proportional to its frequency f, by Planck's formula: E = hf, where h is Planck's constant (6.625 * 10**-34 joule-second). The frequency is inversely proportional to the wavelength w by: f = c/w, where c is the speed of light, 3.0 * 10**8 meters per second. Combine these formulas and we see that the energy is inversely proportional to the wavelength by: E = hc/w If the energy is inversely proportional to the wavelength, a photon with twice the energy has half the wavelength of our 442-nm. photon in this example. So its wavelength is 221 nm. which is in the ultraviolet range.

4 0

3 years ago

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$