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Scilla [17]
2 years ago
13

If my candle has a brightness of 15 candela from a distance of 2 meter. What would the brightness be at a distance of 6 meters?

Physics
1 answer:
nata0808 [166]2 years ago
6 0

Answer:

5 candela

Explanation:

Let the candle brightness be B

Let the distance be d

The candle brightness is inversely proportional to the distance

B α 1/d

B = k/d

If my candle has a brightness of 15 candela from a distance of 2 meter, then;

15 = k/2

k = 30

To get the brightness when the distance is 6m;

B = k/d

B = 30/6

B = 5 candela

Hence the brightness will be 5 candela

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A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m alo
HACTEHA [7]

Answer:

B=2.91\ \mu T

Explanation:

Given that,

The current in the loop, I = 2 A

The radius of the loop, r = 0.4 m

We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}

Put all the values,

B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T

So, the required magnetic field is equal to 2.91\ \mu T.

3 0
2 years ago
URGENT!!!!! If a 6 Ω resistor is in a circuit connected to a voltage source, and the current through the circuit is 2 amps, what
Eduardwww [97]

Answer:

Explanation:

according to ohms law we know that

v=IR

given current =2 amps

given resistance =6Ω

so voltage is

v=2*6 =12 V

3 0
3 years ago
Consider an experiment in which slow neutrons of momentum ¯hk are scattered by a diatomic molecule; suppose that the molecule is
madreJ [45]

Answer:

Check the explanation

Explanation:

When we have an object in periodic motion, the amplitude will be the maximum displacement from equilibrium. Take for example, when there’s a back and forth movement of a pendulum through its equilibrium point (straight down), then swings to a highest distance away from the center. This distance will be represented as the amplitude, A. The full range of the pendulum has a magnitude of 2A.

position = amplitude x sine function(angular frequency x time + phase difference)

x = A sin(ωt + ϕ)

x = displacement (m)

A = amplitude (m)

ω = angular frequency (radians/s)

t = time (s)

ϕ = phase shift (radians)

Kindly check the attached image below to see the step by step explanation to the question above.

3 0
3 years ago
A truck going 15 km􏰀h has a head-on collision with a small car going 30 km􏰀h. Which statement best describes the situation
zlopas [31]

1. e) None of the above is necessarily true.

2.d) Without knowing the mass of the boat and the sack, we cannot tell.

5 0
3 years ago
An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode
sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell v_{0} in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

                v_{x}=v_{0} \times \cos \theta

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of v_{x} and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

           v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}

           v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}

For motion with constant acceleration, we know

            s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}

Along the horizontal, x-axis, we might write this as

            x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}

Measuring distances relative to the firing point means

               x_{0}=0

we know that,

              a_{x}=0

or,

             v_{x}=v_{x 0}=\text { constant }

By applying the values, we get,

           x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}

The acceleration of gravity is vertically downward and is g=-9.8 \mathrm{m} / \mathrm{s}^{2} , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

           y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}

we know, y_{0}=0 and a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}, so,

          y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)

                 y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0
3 years ago
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