A train of 150 m length is going toward north direction at a speed of 10 ms–1 and a parrot is flying towards south direction parallel to the railway track with a speed of 5 ms–1. The time taken by the parrot to cross the train is equal to.?

Answers

Answer 1

Answer:

Answer:

10ms

Explanation:

The bird must travel the length of the train (150m), with a combined speed of 15m/s this means it will take 10s to cross an accumulated 150ms.

Answer 2

Answer:

Answer:

The time taken by the parrot to cross the train is = 10 m/s

Explanation:

given:

A train of 150 m length is going toward north direction at a speed of 10 m/s

and a parrot is flying towards south direction parallel to the railway track with a speed of 5 m/s

find;

The time taken by the parrot to cross the train

using the distance over speed relation formula t = d / v

where:

v = velocity

d = distance

t = time

v = 10 m/s + 5 m/s = 15 m/s (combine velocity)

d = 150 m

t = ?

plugin values into the formula

t = 150

t = 10 m/s

therefore,

The time taken by the parrot to cross the train is = 10 m/s

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A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb

Answers

Answer: 7.578x10^-12

Explanation:

First, we find the power:

Power P = 140x4/100 =5,6W

Distance r = 14m

Then,

Intensity I = P/4πr2

= 5.6/(4π x 14 x 14)

=. 2.27 x10^3 W/m2

Radiation pressure:

P(rad) = I/c =0.00227÷{3 x 10^8)

=7.578x10^-12 N/m2

Answer:

Pr=7.57*10^{11}Pa

Explanation:

We can solve this problem by taking into account the expression

$P_r=(IA)/(c)$

where I is the irradiance, c is the speed of light and A is the area.

We have that the power is 140W, but only 4% is electromagnetic energy, that is

$P=140W=140(J)/(s)\n0.4*140J=56J$

56J is the electromagnetic energy.

The area of the bulb is

$A_b=4\pi r^2=4\pi (14m)^2=2463m^2$

The radiation pressure is

$P_r=(56)/(2463m^2*3*10^8m/s^2)=7.57*10^(-11)Pa$

hope this helps!!

Information that is easily converted into numbers and is stored as a on and off signals is _____ information.

Answers

"Binary" information

How many nanoseconds are in one hour? How do you write the following in scientific notation?2,560,000m

Answers

Answer:

3.6 × 10¹² nanoseconds

Explanation:

Hour is the unit of time. Seconds is the SI unit of time.

Hour and seconds are related as:

1 hour = 60 minutes

1 minute = 60 seconds

So,

1 hour = 60 ×60 seconds = 3600 seconds

Thus,

3600 seconds are in one hour

Also,

1 sec = 10⁹ nanoseconds

Thus,

3600 sec = 3600 × 10⁹ nanoseconds = 3.6 × 10¹² nanoseconds

Thus,

3.6 × 10¹² nanoseconds are in one hour.

Coach ulcer paces the sidelines. Sarting at the 30 yd. line (A), he moves to the 10 yd. line (B), back to the 50 yd. line (C) and finally to the 20yd. Line (D) in 200 seconds. Determine his average speed and velocity.

Answers

Answer:

See the answer below

Explanation:

Average speed = total distance traveled/total time taken

In order to determine the total distance traveled by the coach, consider the attached image.

Distance covered:

30 yd. line to 10 yd. line (A to B)= 20 yds

10 yd. line to 50 yd. line (B to C) = 40 yds

50 yd. line to 20 yd. line (C to D) = 30 yds

Total distance covered = 20 + 40 + 30 = 90 yds

Time taken = 200 seconds

Average Speed = 90/200 = 0.45 yd/s

Velocity = speed with direction

Hence,

His Velocity = 0.45 yd/s to the left of his starting point.

A large rectangular tub is filled to a depth of 2.60 m with olive oil, which has density 915 kg/m3 . If the tub has length 5.00 m and width 3.00 m, calculate (a) the weight of the olive oil, (b) the force of air pressure on the sur- face of the oil, and (c) the pressure exerted upward by the bottom of the tub.

Answers

Answer:

weight is 3.50 x 10^5 N

force is 1.52 * 10^6 N

pressure is 1.25 * 10^5 Pa

Explanation:

given data

Given data

depth = 2.60 m

density = 915 kg/m3

length = 5.00 m

width = 3.00 m

to find out

weight of the olive oil, force of air pressure and the pressure exerted upward

solution

we know density = mass / volume

mass = density* width *length *depth

mass = (915)(3)(5)(2.60)

mass = 3.57 x 10^4 Kg

so weight = mg = 3.57 x 10^4 (9.81) = 3.50 x 10^5 N

weight is 3.50 x 10^5 N

and

we know force = pressure * area

area = 3 *5 = 15 m²

and we know atmospheric Pressure is about 1.01 * 10^5 Pa

so force = 1.01 * 10^5 (15) = 1.52 x 10^6 N

force is 1.52 * 10^6 N

and

we know Fup - Fdown = Weight

Fup = 1.52 * 10^6 + 3.50 * 10^5

Fup = 1.87 * 10^6 N

so pressure = Fup / area

pressure = 1.87 * 10^6 / 15

pressure is 1.25 * 10^5 Pa

In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is recorded on a flat screen-like detector that is 2.00 m away from the slits. If the seventh bright fringe on the detector is 10.0 cm away from the central fringe, what is the wavelength of the light passing through the slits? (The central bright fringe is zeroth one).

Answers

The wavelength of the light passing through the slit is 214 nm.

What is the wavelength?

The wavelength is the distance between identical points in the adjacent cycles of a waveform.

Given that the separation between two slits d is 3.00 × 10^-5 m and the distance from the slit to screen r is 2 m. The distance from the central spot to fringe s is 10.0 m and the bright bands of the spectrum m are 7 for the seventh bright fringe.

The wavelength of the light passing through the slit is calculated as given below.

$\lambda = \frac {sd}{mr}$

$\lambda = \frac {10* 10^(-2) * 3.00 * 10^(-5)}{7 * 2}$

$\lambda = 2.14 * 10^(-7)\;\rm m$

$\lambda = 214 \;\rm nm$

Hence we can conclude that the wavelength of the light passing through the slit is 214 nm.

To know more about the wavelength, follow the link given below.

brainly.com/question/7143261.

To solve this problem, the concepts related to destructive and constructive Interference of light spot and dark spot are necessary.

By definition in the principle of superposition, light interference is defined as

$Y = (m\lambda R)/(d)$

Where,

d = Separation of the two slits

$\lambda = Wavelength$

R = Distance from slit to screen

m= Any integer, which represents the repetition of the spectrum. The order of m equal to 1,2,3,4,5 represent bright bands and the order of m equal to 1.5,2.5,3.5 represent the dark bands.

Y = Distance from central spot to fringe.

Re-arrange the equation to find \lambda we have that

$\lambda = (Yd)/(mR)$