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3 years ago

Outside of the mirror in your bathroom (or bedroom), how else are you using mirrors in your daily life?

Physics

1 answer:

Elan Coil [88]3 years ago

7 0

Answer:

Explanation:

Mirrors are widely used in our day to day life for example

1.E.N.T. specialist used a concave mirror to see the clear and better image of the infected part.

2. A convex mirror is used in vehicles as a rear-view mirror. It covers a large area, so it is commonly used in vehicles.

3. Convex mirrors are generally utilized for making magnifying glasses

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Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us

nalin [4]

Answer:

$13 W/m^2$

Explanation:

The apparent brightness follows an inverse square law, therefore we can write:

$I \propto \frac{1}{r^2}$

where I is the apparent brightness and r is the distance from the Sun.

We can also rewrite the law as

$\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2}$ (1)

where in this problem, we have:

$I_1 = 1300 W/m^2$ apparent brightness at a distance $r_1$ , where

$r_1 = 150$ million km

We want to estimate the apparent brightness at $r_2$ , where $r_2$ is ten times $r_1$ , so

$r_2 = 10 r_1$

Re-arranging eq.(1), we find $I_2$ :

$I_2 = \frac{r_1^2}{r_2^2}I_1 = \frac{r_1^2}{(10r_1)^2}(1300)=\frac{1}{100}(1300)=13 W/m^2$

5 0

2 years ago

Difference between pulling and pushing force

Mrac [35]

Answer:

Push and pull both are forces , but the difference is in their direction at which it is applied . If the force applied in the direction of motion of the particle then we call it as push . If that force applied in the direction OPPOSITE to the motion of particle then it is termed as pull

7 0

2 years ago

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0

3 years ago

How does changes in distance affect the gravitational pull between two objects? Describe and give one example.

maxonik [38]

The formula is

F_grav = G * m1 * m2 / r^2

G m1 and m2 are going to stay the same once chosen no matter what the distance is. The only thing that will change is the distance.

As the distance increases, the Gravitational Force will decrease. It will decrease by quite a bit.

As the distance decreases, the gravitational force will Increase.

The relationship is inverse. The moon travelling around the earth is one example. The earth travelling around the sun is another.

8 0

3 years ago

A regulation basketball has a 32 cm diameter

Rudik [331]

Answer:

1.8 s

Explanation:

Potential energy = kinetic energy + rotational energy

mgh = ½ mv² + ½ Iω²

For a thin spherical shell, I = ⅔ mr².

mgh = ½ mv² + ½ (⅔ mr²) ω²

mgh = ½ mv² + ⅓ mr²ω²

For rolling without slipping, v = ωr.

mgh = ½ mv² + ⅓ mv²

mgh = ⅚ mv²

gh = ⅚ v²

v = √(1.2gh)

v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)

v = 5.47 m/s

The acceleration down the incline is constant, so given:

Δx = 4.8 m

v₀ = 0 m/s

v = 5.47 m/s

Find: t

Δx = ½ (v + v₀) t

t = 2Δx / (v + v₀)

t = 2 (4.8 m) / (5.47 m/s + 0 m/s)

t = 1.76 s

Rounding to two significant figures, it takes 1.8 seconds.

3 0

3 years ago