The table below shows the approximate distances of two stars from Earth.

If 5 mm of rain falls in a 100 m2 field, what volume of rain, in m3, fell in the field?

Nina [5.8K]

The volume of rain that fells in the field is simply given by the area of the field, which is

$A=100 mm^2$

multiplied by the height of rain that fell, which is

$h=5.0 mm$

Therefore, the volume is

$V=hA=(5 mm)(100 mm^2)=500 mm^3$

7 0

4 years ago

The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad

kenny6666 [7]

Answer:

a) v = 2,9992 10⁸ m / s , b) Eo = 375 V / m , B = 1.25 10⁻⁶ T,

c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz , T = 1.05 10⁻¹⁵ s , UV

Explanation:

In this problem they give us the equation of the traveling wave

E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]

a) what the wave velocity

all waves must meet

v = λ f

In this case, because of an electromagnetic wave, the speed must be the speed of light.

k = 2π / λ

λ = 2π / k

λ = 2π / 1.99 10⁷

λ = 3,157 10⁻⁷ m

w = 2π f

f = w / 2 π

f = 5.97 10¹⁵ / 2π

f = 9.50 10¹⁴ Hz

the wave speed is

v = 3,157 10⁻⁷ 9.50 10¹⁴

v = 2,9992 10⁸ m / s

b) The electric field is

Eo = 375 V / m

to find the magnetic field we use

E / B = c

B = E / c

B = 375 / 2,9992 10⁸

B = 1.25 10⁻⁶ T

c) The period is

T = 1 / f

T = 1 / 9.50 10¹⁴

T = 1.05 10⁻¹⁵ s

the wavelength value is

λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm

this wavelength corresponds to the ultraviolet

5 0

4 years ago

A ball of mass 0.6 kg flies through the air at low speed, so that air resistance is negligible. What is the net force acting on

Anettt [7]

To solve this problem we will apply Newton's second law. The second law says that mass by acceleration is equivalent to the Force exerted on a body. Mathematically this

$F = ma$

Here,

m = mass

a = Acceleration

The ball flies through the air under the effect of gravity only. Therefore the acceleration is due to gravity, we have that

$a = - g$

Then,

$F = -mg$

Replacing

$F = -0.6*9.8$

$F = -5.88N$

The negative sign implies that the force acting on the ball is in the downward direction that is towards the negative y direction.

5 0

3 years ago

Which part of the eye controls whether or not the pupil dilates or constricts? iris retina optic nerve cornea

Bas_tet [7]

The answer is Iris (i just took the quiz)

6 0

3 years ago

Read 2 more answers

(a) You short-circuit a 20 volt battery by connecting a short wire from one end of the battery to the other end. If the current

erica [24]

(a) $1.11 \Omega$

When the battery is short-cut, the only resistance in the circuit is the internal resistance of the battery. Therefore, we can apply Ohm's law:

$r=\frac{V}{I}$

where

V = 20 V is the voltage across the internal resistance of the battery

I = 18 A is the current flowing through it

Solving the equation,

$r=\frac{20 V}{18 A}=1.11\Omega$

(b) 360 W

The power generated by the battery is given by the equation

$P=VI$

where

V = 20 V is the voltage of the battery

I = 18 A is the current

Substituting into the formula,

$P=(20 V)(18 A)=360 W$

(c) 360 J

The energy dissipated by the internal resistance is given by

$E=Pt$

where

P = 360 W is the power generated

t = 1 s is the time

Solving the equation, we find

$E=(360 W)(1 s)=360 J$

(d) 1.65 A

The battery is now connected to a $R=11 \Omega$ resistor. This means that the internal resistance of the battery is now connected in series with the other resistor R: so, the total resistance of the circuit is

$R_T = r+R=1.11 \Omega +11 \Omega = 12.11 \Omega$

And so, the current flowing through the circuit is

$I=\frac{V}{R_T}=\frac{20 V}{12.11\Omega}=1.65 A$

(e) 29.9 W

The power dissipated in the external resistor is given by

$P=I^2 R$

where

I = 1.65 A is the current

$R=11 \Omega$ is the resistance

Solving the equation, we find

$P=(1.65 A)^2(11 \Omega)=29.9 W$

(f) 18.17 V

The terminals of the voltmeter are placed at the two end of the battery. The battery provides an emf of 20 V, however due to the internal resistance, some of this voltage is dropped across the internal resistance. Therefore, the actual potential difference that will be read by the voltmeter will be:

$V=\epsilon - Ir =20 V -(1.65 A)(1.11 \Omega)=18.17 V$

4 0

3 years ago