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Ainat [17]
2 years ago
10

1.A virtual image is formed 20.5 cm from a concave mirror having a radius of curvature of 31.5 cm.

Physics
1 answer:
Mashutka [201]2 years ago
4 0

The position of the object is = -68cm

The magnification of the mirror= 0.3

<h3>Calculation of object distance</h3>

The image distance = 20.5cm

The focal length= R/2 = 31.5/2= 15.75

The object distance= ?

Using the lens formula,1/f = 1/v-1/u

1/u = 1/v- 1/f

1/u = 1/20.5 - 1/15.75

1/u = 0.0489- 0.0635

1/u = -0.0146

u = -68cm

The magnification of the mirror is image size/object size

= 20.5cm/-68cm

= 0.3

Learn more about lens here:

brainly.com/question/9757866

#SPJ1

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Velocity of a particle varies with its displacement as v = ( √(9 ... Velocity of a particle varies with its displacement as v = ( √(9 - x^2) ) m/sFind the magnitude of maximum acceleration of the particle.

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Effects of the earths orbit around the sun
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7 0
3 years ago
A crane exerts a net force of 900 N upward on a 750-kilogram car as the crane starts to lift the car from the deck of a cargo sh
melisa1 [442]

Answer: 1.2m/s^2

Explanation: the force exerted on the car is 900N upwards

The mass of the car is 750kg

According to Newton's third law acceleration is proportional to force

F = ma

900 = 750a

a = 900/750

a = 1.2m/s^2

5 0
4 years ago
In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
muminat

Answer:

2.1\cdot 10^{21} electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

E=\frac{kQ}{r^2}

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a  charged sphere, so we have:

E_b=3.00\cdot 10^6 N/C is the maximum electric field strength tolerated by the air before breakdown occurs

r=1.00 km = 1000 m is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C

Assuming that the cloud is negatively charged, then

Q=-333.3 C

And since the charge of one electron is

e=-1.6\cdot 10^{-19}C

The number of excess electrons on the cloud is

N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}

5 0
3 years ago
A 10.00 kg mass is moving to the right with a velocity of 14.0 m/s. A 12.0 kg mass is moving to the left with a velocity of 8.00
Basile [38]

Answer:

2 m/s

Explanation:

From the conservation of momentum, the initial momentum of the system must be equal to the final momentum of the system.

Let the 10.00 kg mass be m_1 and the 12.0 kg mass be m_2. When they collide and stick, they have a combined mass of m_1+m_2.

Momentum is given by p=mv. Set up the following equation:

\displaystyle m_1v_1+m_2v_2=(m_1+m_2)v_f, where v_f is the desired final velocity of the masses.

Call the right direction positive. To indicate the 12.0 kg object is travelling left, its velocity should be substitute as -8.00 m/s.

Solving yields:

10\cdot 14 + 12\cdot (-8)=(10+12)v_f\\\implies v_f=\boxed{2 \text{ m/s}}

4 0
2 years ago
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