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3 years ago

A man walks backward on a bus. To a stationary observer, the bus moves

Physics

1 answer:

NNADVOKAT [17]3 years ago

6 0

Explanation:

V(m,b)= V(m,e) - V(b,e)

2.3m/s (west) = V(m,e) - 15m/s (east)

to solve this change the direction of slower speed. so it will become negative value

-2.3m/s ( east) = V (m,e) - 15m/s (east)

V(m,e) = 15m/s (east) - 2.3m/s (east)

V(m,e) = 12.7 m/s (east)

velocity of the man in frame of reference of stationary observer is 12.7 m/s

it's slightly less than the velocity of bus so stationary observer see the man move slower than the one inside the bus

ps:- V(m,b) - velocity of man relative to bus

V(b,e) - velocity of bus relative to earth

V(m,e) - velocity of man relative to earth

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Given the wave function: Y(x,t) = 5sin27(0.2x - 3t); (x = meters, t = sec.): What are the amplitude, frequency, wavelength, angu

MariettaO [177]

Answer:

Explanation:

Y = 5 Sin27( .2x-3t)

= 5 Sin(5.4x - 81 t )

Amplitude = 5 m

Angular frequency ω = 81

frequency = ω / 2π

= 81 / (2 x 3.14 )

=12.89

Wave length λ = 2π / k ,

k = 5.4

λ = 2π / 5.4

= 1.163 m

Phase velocity =ω / k

= 81 / 5.4

15 m / s.

The wave is travelling in + ve x - direction.

3 0

3 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

3 years ago

a large heavy truck and a baby carriage roll down a hill. Neglecting friction, at the bottom of the hill, the baby carriage will

Lisa [10]

Answer:

please give me brainlist and follow

Explanation:

At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.

5 0

3 years ago

Somebody can tell me what answers it is?

sergij07 [2.7K]

Answer:

Explanation:

The color you see is the color the object reflectes. The rest of the color are absorbed by that object.

6 0

3 years ago

Read 2 more answers

Someone help please by providing work and answers please :)

Nastasia [14]

First we gotta use an equation of motion:

$d = ut + \frac{1}{2} a {t}^{2}$

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

$100 = 0 + \frac{1}{2} (9.8) {t}^{2}$

Now we just need to solve for t:

${t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}$

Hit the calculators, and you'll get 4.5 seconds!

5 0

3 years ago