What are earths two main motions called

A low resistance light bulb and a high resistance light bulb are connected in parallel with each other. Which bulb is brighter i

sweet [91]

<h2>Answer:</h2>

The bulb with low resistance will be brighter.

<h2>Explanation:</h2>

The brightness of a bulb is a function of both the voltage across the bulb and current flowing through the bulb. The higher the voltage, the higher the current. Hence the brighter the bulb.

Now, according to the question, the bulbs (the high resistance bulb and the low resistance bulb) are connected in parallel with each other. This means that the same voltage passes across them.

Also, we know that according to Ohm's law, the voltage (V) and current (I) through a conductor are related by the following equation;

V = I x R -------------------(i)

Where;

R is the resistance of the conductor.

We can re-write equation (i) as follows;

I = V / R -----------------------(ii)

According to equation (ii), at fixed voltage (V), the current (I) will increase as the resistance (R) decreases.

Now, since the two bulbs have the same voltage, the bulb with the low resistance will allow a larger flow of current than the bulb with high resistance. Therefore, as said earlier that brightness is dependent on voltage and current, the bulb with the low resistance (and having larger current at some voltage) will be brighter than the bulb with the high resistance (having smaller current at same voltage).

6 0

3 years ago

The three types of nuclear radiation<br>

Eva8 [605]

The three types are alpha beta and gamma

6 0

3 years ago

The displacement of a wave traveling in the negative y-direction is D(y,t) = ( 4.60 cm ) sin ( 6.20 y+ 60.0 t ), where y is in m

trapecia [35]

Answer:

The question is incomplete, below is the complete question

"The displacement of a wave traveling in the negative y-direction is D(y,t) = ( 4.60cm ) sin ( 6.20 y+ 60.0 t ), where y is in m and t is in s.

A) What is the frequency of this wave?

B) What is the wavelength of this wave?

C) What is the speed of this wave?"

Answers:

a. $f=\frac{30}{\pi }Hz\\$

b. $wavelength=\frac{\pi }{3.1}m \\$

c. $v=9.68m/s$

Explanation:

The equation of a wave is represented as

$D(x,t)=Asin(kx+wt) \\$

Where A=amplitude

w=angular frequency=2πf

K=wave numbers =2π/λ

since we re giving he equation D(y,t) = ( 4.60cm ) sin ( 6.20 y+ 60.0 t ),

we can compare and get the value for the wave number and angular frequency.

By comparing we have

w=60rads/s

k=6.20

a. to determine the frequency, from the expression fr angular wave frequency we have

w=2πf hence

f=w/2π

if we substitute we arrive at

$f=\frac{60}{2\pi }\\f=\frac{30}{\pi }Hz\\$

b. to determine the wave length, we use

$k=\frac{2\pi }{wavelength} \\k=6.2\\wavelength=\frac{2\pi }{k} \\wavelength=\frac{2\pi }{6.2} \\wavelength=\frac{\pi }{3.1}m \\$

c. the wave speed v is express as the product of the frequency and the wavelength. Hence

$v=frequency*wavelength \\v=\frac{30}{\pi } *\frac{\pi }{3.1}\\ v=9.68m/s$

6 0

3 years ago

Using a power supply, I place +1.00µC on one plate of a capacitor. A capacitor is a device for storing charge. I, then, connect

Paraphin [41]

Answer:

$N=62421972534 electrons$

Explanation:

The charge transferred charge Q= −10nC = -10*10-9 C

Charge of a electron : $Q_{e}= -1,602*10^{-19}C$

Number of electrons transferred: $N=\frac{Q}{Q_{e} }=(-10*10^{-9} )/(-1,602*10^{-19})=62421972534 electrons$

5 0

3 years ago

2. A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.5 . At 30.0 s after blast

harina [27]

Answer:

a)The highest point reached by the rocket is 1412 m

b)The rocket crashes after 54.7 s

Explanation:

Hi there!

The equations of height and velocity of the rocket are the following:

h = h0 + v0 · t + 1/2 · a · t² (while the engines work).

h = h0 + v0 · t + 1/2 · g · t² (when the rocket is in free fall).

v = v0 + a · t (while the engines work).

v = v0 + g · t (when the rocket is in free fall).

Where:

h = height of the rocket at a time t.

h0 = initial height of the rocket.

v0 = initial velocity.

t = time.

a = acceleration due to the engines.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the rocket at a time t.

First, let's find the velocity and height reached by the rocket until the engines fail:

h = h0 + v0 · t + 1/2 · a · t²

Let's set the origin of the frame of reference at the launching point so that h0 = 0. Since the rocket starts from rest, v0 = 0. So after 30.0 s the height of the rocket will be:

h = 1/2 · a · t²

h = 1/2 · 2.5 m/s² · (30.0 s)²

h = 1125 m

Now let's find the velocity of the rocket at t = 30.0 s:

v = v0 + a · t (v0 = 0)

v = 2.5 m/s² · 30.0 s

v = 75 m/s

After 30.0s the rocket will continue to ascend with a velocity of 75 m/s. This velocity will be gradually reduced due to the acceleration of gravity. When the velocity is zero, the rocket will start to fall. At that time, the rocket is at its maximum height. So, let's find the time at which the velocity of the rocket is zero:

v = v0 + g · t

0 = 75 m/s - 9.8 m/s² · t (v0 = 75 m/s because the rocket begins its free-fall motion with that velocity).

-75 m/s / -9.8 m/s² = t

t = 7.7 s

Now, let's find the height of the rocket 7.7 s after the engines fail:

h = h0 + v0 · t + 1/2 · g · t²

The rocket begins its free fall at a height of 1125 m and with a velocity 75 m/s, then, h0 = 1125 m and v0 = 75 m/s:

h = 1125 m + 75 m/s · 7.7 s - 1/2 · 9.8 m/s² · (7.7 s)²

h = 1412 m

The highest point reached by the rocket is 1412 m

b) Now, let's calculate how much time it takes the rocket to reach a height of zero (i.e. to crash) from a height of 1412 m.

h = h0 + v0 · t + 1/2 · g · t² (v0 = 0 because at the maximum height the velocity is zero)

0 = 1412 m - 1/2 · 9.8 m/s² · t²

-1412 m / -4.9 m/s² = t²

t = 17 s

The rocket goes up for 30.0 s with an acceleration of 2.5 m/s².

Then, it goes up for 7.7 s with an acceleration of -9.8 m/s².

Finally, the rocket falls for 17 s with an acceleration of -9.8 m/s²

The rocket crashes after (30.0 s + 7.7 s + 17 s) 54.7 s

6 0

3 years ago