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3 years ago

an object is placed in front of a curved mirror as shown. which of the labeled positions is the current position of the image

Physics

2 answers:

alexandr402 [8]3 years ago

4 0

Answer:A

Explanation:

The image is formed behind the mirror.

Lilit [14]3 years ago

4 0

Answer:

Explanation:

took the test

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The discovery of Uranus________.

Yuki888 [10]

I think disproved geocentric theory proved that planetary orbits are elleptical

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3 years ago

Read 2 more answers

A ball with a mass of 3kg is dropped from the top of a building this is 20m high. what is the velocity of the ball when it is 10

anzhelika [568]

Answer:

Velocity=14[m/s]

Explanation:

We can solve this problem by using the principle of energy conservation, where potential energy becomes kinetic energy.

In the attached image we can see the illustration of the ball falling from the height of 20 meters, at this time the potential energy will have the following value.

$Ep=m*g*h\\where:\\m=3[kg]\\h=20[m]\\$

$Ep=3*9.81*20\\Ep=588.6[J]$

When the ball passes through half of the distance (10m) its potential energy will have decreased by half as shown below.

$Ep=3*9.81*10\\Ep=294.3[m]$

If we know that potential energy is transformed into kinetic energy, we can find the value of speed.

$Ek=\frac{1}{2} *m*v^{2} \\therefore\\v=\sqrt{\frac{Ek*2}{m} } \\v=\sqrt{\frac{294.3*2}{3} } \\\\v=14[m/s]$

7 0

3 years ago

A car is traveling around a horizontal circular track with radius r = 210 m at a constant speed v = 23 m/s as shown. The angle θ

Fantom [35]

Centripetal acceleration of car is given by formula

$a_c = \frac{v^2}{R}$

now plug in the values in this

$a_c = \frac{23^2}{210}$

$a_c = 2.52 m/s^2$

Part b)

At position A we have

x component of acceleration is given as

$a_x = -a_c cos23$

$a_x = -2.32 m/s^2$

Part c)

At position A we have

y component of acceleration is given as

$a_y = -a_c sin23$

$a_y = -0.98 m/s^2$

Part d)

At position B we have

x component of acceleration is given as

$a_x = -a_c cos53$

$a_x = -1.52 m/s^2$

Part e)

At position B we have

Y component of acceleration is given as

$a_y = a_c sin53$

$a_y = 2.01 m/s^2$

6 0

4 years ago

A sprinter in a 100-m race accelerates uniformly for the first 71 m and then runs with constant velocity. The sprinter’s time fo

adoni [48]

Answer:

The acceleration of the sprinter is 1.4 m/s²

Explanation:

Hi there!

The equation of position of the sprinter is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the sprinter at a time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

Since the origin of the frame of reference is located at the starting point and the sprinter starts from rest, then, x0 and v0 are equal to zero:

x = 1/2 · a · t²

At t = 9.9 s, x = 71 m

71 m = 1/2 · a · (9.9 s)²

2 · 71 m / (9.9 s)² = a

a = 1.4 m/s²

The acceleration of the sprinter is 1.4 m/s²

3 0

3 years ago

Monochromatic light of variable wavelength is incident normally on a thin sheet of plastic film in air. The reflected light is a

BartSMP [9]

Answer:

thickness t = 528.433 nm

Explanation:

given data

wavelength λ1 = 477.1 nm

wavelength λ2 = 668.0 nm

n = 1.58

solution

we know for constructive interference condition will be

2 × t × μ = (m1+0.5) × λ1 ....................1

2 × t × μ = (m2+0.5) × λ2 ....................2

so we can say from equation 1 and 2

(m1+0.5) × λ1 = (m2+0.5) × λ2

$\frac{\lambda 2}{\lambda 1} = \frac{m1+0.5}{m2+0.5}$ ..............3

put here value and we get

$\frac{668.0}{477.1} = \frac{m1+0.5}{m2+0.5}$

$\frac{m1+0.5}{m2+0.5}$ = 1.4

$\frac{m1+0.5}{m2+0.5} = \frac{7}{5}$ ...................4

so we here from equation 4

m1+0.5 = 7

m1 = 3 .................5

m2+0.5 = 4

m2 = 2 .................6

so now put value in equation 1

2 × t × μ = (m1+0.5) × λ1

2 × t × 1.58 = (3+0.5) × 477.1

solve it we get

thickness t = 528.433 nm

4 0

3 years ago