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3 years ago

Why do we see ourselves in the mirror?

Physics

2 answers:

Crazy boy [7]3 years ago

5 0

Answer:

When the light hits the surface on the mirror, some of the light bounces back and you get your reflection.

Arlecino [84]3 years ago

3 0

You can see yourself in a mirror because light rays<span> bounce off its shiny surface. Light rays come from everything you can see, including yourself. You see things when </span>the light<span>rays from them enter yours eyes. Some of the light rays that come from yourself strike the mirror. The mirror reflects the rays because it is very smooth. The rays come back to your and enter your eyes.</span>

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a car has a velocity of 10ms-1. it accelerates at 0.2 ms-2 for half minute find the total distance travelled and final velocity

Readme [11.4K]

Answer:

Final velocity = 16 m/s

Total distane travelled = 390 m

Explanation:

We can use equation of motion to solve this:

$v = u + at \\ v = 10 + 0.2(30) \\ v = 16 {ms}^{ - 1}$

$s = ut + \frac{1}{2} a {t}^{2} \\ s = 10(30) + \frac{1}{2} (0.2) {(30)}^{2} \\ s = 390m$

5 0

3 years ago

A 2.5 kg ball rolls forward at 10.0 m/s. What is the ball's momentum?

Anuta_ua [19.1K]

Answer:25kgm/s

Explanation:

mass=2.5kg

Velocity=10m/s

Momentum=mass x velocity

Momentum=2.5 x 10

Momentum=25kgm/s

4 0

4 years ago

A circular loop of radius r carries a current i. at what distance along the axis of the loop is the magnetic field one-half its

lana [24]

At r = 0.766 R the magnetic field intensity will be half of its value at the center of the current carrying loop.

We have a circular loop of radius ' r ' carrying current ' i '.

We have to find at what distance along the axis of the loop is the magnetic field one-half its value at the center of the loop.

<h3>What is the formula to calculate the Magnetic field intensity due to a current carrying circular loop at a point on its axis?</h3>

The formula to calculate the magnetic field intensity due to a current carrying ( i ) circular loop of radius ' R ' at a distance ' x ' on its axis is given by -

$B(x) = \frac{\mu_{o} iR^{2} }{2(x^{2} +R^{2})^{\frac{3}{2} } }$

Now, for magnetic field intensity at the center of the loop can calculated by putting x = 0 in the above equation. On solving, we get -

$B(0) = \frac{\mu_{o} i}{2R}$

Let us assume that the distance at which the magnetic field intensity is one-half its value at the center of the loop be ' r '. Then -

$\frac{\mu_{o} iR^{2} }{2(r^{2} +R^{2})^{\frac{3}{2} } } = \frac{1}{2} \frac{\mu_{o}i }{2R}$

$2R^{3} = (r^{2} +R^{2} )^{\frac{3}{2} }$

$4R^{6} = (r^{2} +R^{2} )^{3}$

$r^{2} =0.587R^{2}$

r = 0.766R

Hence, at r = 0.766 R - the magnetic field intensity will be half of its value at the center of the current carrying loop.

To solve more questions on magnetic field intensity, visit the link below-

brainly.com/question/15553675

#SPJ4

6 0

2 years ago

3. Label the parts of each wave.<br> Please help me!!!

lisabon 2012 [21]

Answer:

D is the wavelength

A is the crest

C is amplitude

B is trough

G is wavelength

H is compression

I is rarefaction

4 0

3 years ago

Lubov Fominskaja [6]

Answer:

1.5 hr

16.7

Explanation:

Zero apparent weight means there's no normal force.

Sum the forces in the centripetal direction.

∑F = ma

mg = mv²/r

v = √(gr)

v = √(7.4×10⁶ m × 10 m/s²)

v = 8602 m/s

The circumference of the equator is:

C = 2πr

C = 2π (7.4×10⁶ m)

C = 4.65×10⁷ m

So the period is:

T = C / v

T = (4.65×10⁷ m) / (8602 m/s)

T = 5405 s

T = 1.5 hr

The initial speed is:

v = C / T

v = (4.65×10⁷ m) / (25 h × 3600 s/h)

v = 517 m/s

The speed increases by a factor of:

8602 m/s / 517 m/s

16.7

3 0

3 years ago