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Setler79 [48]
3 years ago
7

Imagine you leave the dock on a fishing boat that travels east for 30 kilometers when the captain realizes he left the fishing r

ods on the dock. It takes him 40 minutes to travel out against the current but only 20 minutes to return to the dock.
Physics
1 answer:
telo118 [61]3 years ago
3 0

Answer:

Average speed is 60 km/hr whereas average velocity is 0 km/ hr.

Explanation:

The average speed of the boat is 60 km/ hr while on the other hand, the average velocity of the boat is 0 km/ hr because average speed is the total distance covered by the boat in total time and average velocity is the displacement covered by the boat in total time. The total distance covered by the boat is 60 km and the total time is 60 minutes which is equals to one hour so the answer is 60 km/hr whereas the displacement is the shortest distance between initial and final position which is 0 in this case so 0 divided by 60 minutes or one hour is also 0.

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3 years ago
Dont use me for points please i need help, (circuits for physics)
zhannawk [14.2K]

Answer:

B, A, A, B

Explanation:

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2 years ago
A cubical box measuring 1.29 m on each side contains a monatomic ideal gas at a pressure of 2.0 atm How much thermal energy do t
Marrrta [24]

Answer:

a) U = 652.545\,kJ, b) v \approx 659.568\,\frac{m}{s}

Explanation:

a) According to the First Law of Thermodinamics, the system is not reporting any work, mass or heat interactions. Besides, let consider that such box is rigid and, therefore, heat contained inside is the consequence of internal energy.

Q = U

The internal energy for a monoatomic ideal gas is:

U = \frac{3}{2} \cdot n \cdot R_{u} \cdot T

Let assume that cubical box contains just one kilomole of monoatomic gas. Then, the temperature is determined from the Equation of State for Ideal Gases:

T = \frac{P\cdot V}{n\cdot R_{u}}

T = \frac{(202.65\,kPa)\cdot(1.29\,m)^{3}}{(1\,kmole)\cdot(8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K} )}

T = 52.325\,K

The thermal energy contained by the gas is:

U = \frac{3}{2}\cdot (1\,kmole)\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K})\cdot (52.325\,K)

U = 652.545\,kJ

b) The physical model for the cat is constructed from Work-Energy Theorem:

U = \frac{1}{2}\cdot m_{cat} \cdot v^{2}

The speed of the cat is obtained by isolating the respective variable and the replacement of every known variable by numerical values:

v = \sqrt{\frac{2 \cdot U}{m_{cat}}}

v = \sqrt{\frac{2\cdot (652.545 \times 10^{3}\,J)}{3\,kg} }

v \approx 659.568\,\frac{m}{s}

3 0
3 years ago
If the distance between slits on a diffraction grating is 0.50 mm and one of the angles of diffraction is 0.25°, how large is th
Trava [24]

Answer:

-  path differnce = 2.18*10^-6

-  1538 lines

Explanation:

- The path difference for the waves that produce the pattern of diffraction, is given by the following formula:

path\ difference\ = dsin\theta           (1)

d: separation between slits = 0.50mm = 0.50*10^-3 m

θ: angle of a diffraction = 0.25°

Then, the path difference is:

path\ difference\ =(0.50*10^{-3}m)sin(0.25\°)=2.18*10^{-6}m

- The maximum number of bright lines are calculated by using the following formula:

m\lambda = dsin\theta           (2)

m: order of the bright

λ: wavelength = 650nm

The maximum bright is calculated for an angle of 90°:

m=\frac{(0.50*10^{-3}m)sin90\°}{650*10^{-9}m} \approx 769

The maxium number of bright lines are twice the previous result, that is, 1538 lines

8 0
3 years ago
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