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Shtirlitz [24]
2 years ago
13

A spiracle polish metallic ball is uses a diverging mirror with a focal length of -20 cm a bird 25 cm tall standing 50 cm away L

ooks directly at the mirror what are the size and position of the birds image
Physics
1 answer:
Aleksandr-060686 [28]2 years ago
5 0

Answer:

1/f = 1/i + 1/o      thin lens equation

1/i = 1/f - 1/o

i =  o * f / ( o - f) = 50 * (-20) / (50 - (-20)) = -14.3 cm

The final image is  erect and 14.3 cm behind the curved surface

M = -o / i = 14.3 / 50 = .29      magnificaton of object

S = .29 * 25 cm = 7.1 cm       appearance of bird in mirror (height)

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Las 3 leyes de klepler del sistema planetario<br><br><br><br> AYUDAA¡¡
shtirl [24]
I actually don't know. i speak english.
5 0
3 years ago
Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma
Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

Mass of inertia is given by

I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2

Angular acceleration is given by

\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454

The coefficient of friction is 0.54454

At r = 0.25 m

\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N

The force needed to stop the wheel is 104.00902 N

5 0
3 years ago
One end of a horizontal spring with force constant 130.0 Ni'm is attached to a vertical wall. A 4.00-kg block sitting on the flo
tekilochka [14]

Answer:

Explanation:

When the spring is compressed by .80 m , restoring force by spring on block

= 130 x .80

= 104 N , acting away from wall

External force = 82 N , acting towards wall

Force of friction acting towards wall = μmg

= .4 x 4 x 9.8

= 15.68 N

Net force away from wall

= 104 -15.68 - 82

= 6.32 N

Acceleration

= 6.32 / 4

= 1.58 m / s²

It will be away from wall

Energy released by compressed spring = 1/2 k x²

= .5 x 130 x .8²

= 41.6 J

Energy lost in friction

= μmg x  .8

= .4 x 4 x 9.8 x .8

= 12.544 J

Energy available to block

= 41.6 - 12.544 J

= 29 J

Kinetic energy of block = 29

1/2 x 4 x v² = 29

v = 3.8 m / s

This will b speed of block as soon as spring relaxes. (x = 0 )

4 0
3 years ago
Where is the epicenter of the hypothetical earthquake as shown in the illustration below?
Genrish500 [490]

Answer:

Point D

Explanation:

The epicenter of a hypothetical earthquake is located at the point where the earthquake begins.

(See the attached image).

Hope it helps!

5 0
3 years ago
What two factors will make an object stable ​
galben [10]

The position of the centre of gravity of an object affects its stability. The lower the centre of gravity (G) is, the more stable the object. The higher it is the more likely the object is to topple over if it is pushed. Racing cars have really low centres of gravity so that they can corner rapidly without turning over.

Increasing the area of the base will also increase the stability of an object, the bigger the area the more stable the object. Rugby players will stand with their feet well apart if they are standing and expect to be tackled.

8 0
3 years ago
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