All categories

4 years ago

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

Physics

1 answer:

NNADVOKAT [17]4 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

You might be interested in

Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh

MariettaO [177]

Answer:

The separation between the two lowest levels = $1.24 * 10^{-39}J$

The values of n where the energy of molecule reaches 1/2 kT at 300K = $2.2 * 10^{9}$

The separation at this level = 1.8 * $10^{-30}$ J

Explanation:

Knowing the formula

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$

Mass of oxygen molecule

m (O2) = 32 amu * $\frac{1.6605 * 10^{-27 kg} }{1 amu}$

So the energy diference between the two lowest levels:

E2 - E1 = $\frac{3h^{2} }{8mL^{2} }$

E2 - E1 = $\frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2} } = 1.24 * 10^{-39}J$

Now we should find n where the energy of molecule reaches 1/2 kT

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$ = $\frac{1}{2}kT$

$\frac{h^{2} }{8 mL^{2} }$ = $4.13 * 10^{-14}J$

$n^{2} * (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K$

$n = 2.2 * 10^{9}$

by the end is necessary to calculate the separation of the level

En - En-1 = $(n^{2} - (n - 1)^{2}) * \frac{h^{2} }{8 mL^{2} }$

= 1.8 * $10^{-30}$ J

4 0

3 years ago

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts help.

GrogVix [38]

Answer:

$H = 532 m$

Explanation:

When teacher falls into the cliff then she shout for Help at the same time

so here we know that sound will go down and reflect back up

so here in 3 s distance traveled by the sound

$d = vt$

$d = 340 \times 3$

$d = 1020 m$

now in the same time the distance that teacher will fall down is given as

$d_1 = \frac{1}{2}gt^2$

$d_1 = \frac{1}{2}(9.81)(3^2)$

$d_1 = 44.1 m$

now total distance traveled by teacher and sound in 3 s

$d_{total} = d + d_1$

$d_{total} = 1020 + 44.1$

this total distance must be equal to twice the height of the cliff

$2H = 1064.1 m$

$H = 532 m$

7 0

4 years ago

Read 2 more answers

If jack was traveling north 120 miles and it took him 3 hours to get there. what is the velocity that jack was traveling? is thi

Dimas [21]

Answers:

40 mp/h; Vector

Reason:

120/3 is 40 miles per hour.

Velocity is a vector measurement.

^.^

- Amanda

4 0

3 years ago

Why would you have trouble breathing at high altitudes?

monitta

Answer:

A. It is colder at the top of a mountain

Explanation:

4 0

3 years ago

.A box falls to the ground from a delivery truck traveling at 30 m/s. After hitting the road, it slides 45 m to

Romashka-Z-Leto [24]

Answer:

t = 3 seconds

Explanation:

Given that,

Initial speed, u = 30 m/s

Final speed, v = 0

It slides 45 m to rest.it take the box to come to rest

We need to find how long it take the box to come to rest.

Let a be the acceleration and t is time.

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(30)^2-u^2}{2(45)}\\\\=10\ m/s^2$

Now finding time.

$t=\dfrac{v-u}{a}\\\\t=\dfrac{30-0}{10}\\\\t=3\ s$

So, the required time is 3 seconds.

8 0

3 years ago