If you walk 3 kilometers in 30 minutes , what is the average speed in kilometers per hour?

A 10 kg migratory swan cruises at 20m/s. A calculation that takes into ac-count the necessary forces shows that this motion requ

IgorC [24]

Answer:

Part A:

Distance=864000 m=864 km

Part B:

Energy Used=ΔE=8638000 Joules

Part C:

$\frac{\triangle m}{m}=0.004998=0.49985\%$

Explanation:

Given Data:

v=20m/s

Time =t=12 hours

In Secs:

Time=12*60*60=43200 secs

Solution:

Part A:

Distance = Speed**Time

Distance=v*t

Distance= 20*43200

Distance=864000 m=864 km

Part B:

Energy Used=ΔE= Energy Required-Kinetic Energy of swans

Energy Required to move= Power Required*time

Energy Required to move=200*43200=8640000 Joules

Kinetic Energy= $\frac{1}{2}mv^2$

$K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules$

Energy Used=ΔE=8640000 -2000

Energy Used=ΔE=8638000 Joules

Part C:

Fraction of Mass used=Δm/m

For This first calculate fraction of energy used:

Fraction of energy=ΔE/Energy required to move

ΔE is calculated in part B

Fraction of energy=8638000/8640000

Fraction of energy=0.99977

Kinetic Energy= $\frac{1}{2}mv^2$

Now, the relation between energies ratio and masses is:

$\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2$

$\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977$

$\frac{\triangle m}{m}=0.004998=0.49985\%$

3 0

2 years ago

What is the acceleration of an 24 kg object that applies a force of 130N?

WITCHER [35]

Answer:

To find the acceleration of the object we have to apply Newton second law of motion that is F = mass × acceleration.

Explanation:

Given ,

F = 130N

M = 24kg

A = ?

F = m× a

then ,

130N = 24kg ×a

a = 130/24 = 5 m/s.

6 0

1 year ago

Radio waves travel at 3.00 · 108 m/s. Calculate the wavelength of a radio wave of frequency 900 kHz. (9.00 · 105 Hz.)

Anuta_ua [19.1K]

V: velocity of wave
f: frequency
L: wavelenght

v = fL => L = v/f => L = (3x10^8)/(900x10^3) => L = 3.33 x 10^2m

7 0

2 years ago

A beam of light strikes one face of a window pane with an angle of incidence of 30.0°. The index of refraction of the glass is 1

natka813 [3]

Answer:

(a). The angle of refraction is 19.26°.

(b). That is proved that the rays in air on either side of the glass are parallel to each other

Explanation:

Given that,

Angle of incidence = 30.0°

Index of reflection of glass = 1.52

(a). We need to calculate the angle of refraction for the ray inside the glass

Using snell's law

$\dfrac{\sin i}{\sin r}=\mu$

$\sin r=\dfrac{\sin i}{\mu}$

Put the value into the formula

$\sin r=\dfrac{\sin 30}{1.52}$

$r=\sin^{-1}(0.329)$

$r=19.26^{\circ}$

(b). We know that,

The incident ray and emerging ray is equal then the ray will be parallel.

We need to prove that the rays in air on either side of the glass are parallel to each other

Using formula for emerging ray

$\dfrac{\sin e}{\sin r}=\mu$

$\sin e=\sin r\times \mu$

Put the value into the formula

$\sin e=0.3289\times 1.52$

$e=\sin^{-1}(0.499)$

$e=29.9\approx 30^{\circ}$

So, $\sin i=\sin e$

This is proved.

Hence, (a). The angle of refraction is 19.26°.

(b). That is proved that the rays in air on either side of the glass are parallel to each other

6 0

3 years ago

Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o

IrinaVladis [17]

Answer:

The net gravitational force on the mass is $1.27\times 10^{-5}$

Explanation:

We have by Newton's law of gravity the force of attraction between masses $m_{1},m_{2}$

$F_{att}=G\frac{m_{1}m_{2}}{r^{2}}$

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

$F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N$

Force of attraction between 435 kg mass and 38 kg mass is

$F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N$

Thus the net force on mass 38.0 kg is $F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}$

8 0

3 years ago