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fredd [130]
3 years ago
15

A concave mirror is used as a reflector ina torch light?

Physics
1 answer:
4vir4ik [10]3 years ago
8 0

Answer:

A concave mirror is used as a torch reflector. ... When a light bulb is placed at the focus of a concave mirror reflector, the diverging light rays of the bulb are collected by the reflector. These rays are then reflected to produce a strong, parallel-sided beam of light.

Explanation:

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Class characteristics serve as corroboration for other, more subjective pieces of evidence in a courtroom (like witness testimon
Amiraneli [1.4K]

Answer: True

Explanation:

Class characteristics can be define as the features which are common to the group of objects. Like the make, model, label of the manufacturing company, design, shape and form. The individual characteristics can be define as the features which develop on the object or any other article with it's wear and use. Like tear, cuts, malformation and deposition of dust, dirt, and mud. The individual characteristic indicate towards the ownership of article or evidence to a particular person.

The class characteristics can only support the possibility of the evidence exactly alike that of the evidence found at the scene of crime. But the individual characteristics can directly link the evidence with the cause of crime. Hence, will be useful to prove that a crime has taken place in the court of law.

8 0
3 years ago
Read 2 more answers
The brightness of a star is determined
nasty-shy [4]
100% C . By size and distance
4 0
2 years ago
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The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick
Travka [436]

Answer:

c. vf is greator than v2, but less than v1

Explanation:

The principle of conservation of linear momentum states that when two or more bodies act upon one another, their total momentum remains constant.

In a system of colliding bodies the total momentum of the system just before the collision is the same as the total momentum just after the collision.  

Collisions in which the kinetic energy is conserved are called elastic collision.

Collisions in which the kinetic energy is not conserved are called inelastic collisions.  If the two objects stick together after the collision and move with a common velocity, the collision is said to be perfectly inelastic.

<em>The above scenario is a perfectly inelastic collision. The initial velocity of particle 1 was greater than particle 2 before collision. After collision, its velocity will reduce to a final velocity vf as it transfers some of its kinetic energy to particle 2; whereas, the velocity of particle 2 will increase to a final velocity vf as it absorbs some of the kinetic energy of particle 1.</em>

Therefore,

a. vf = v2 is wrong because vf is greater than v2

b. vf is less than v2 is wrong because vf is greater than v2

c. vf is greater than v2, but less than v1 is correct.

d. vf = v1 is wrong because vf is less than v1

4 0
3 years ago
A child is riding a merry-go-round that has an instantaneous angular speed of 12 rpm. If a constant friction torque of 12.5 Nm i
sammy [17]

Answer:

-0.25 rad/s^2

Explanation:

The equivalent of Newton's second law for rotational motions is:

\tau = I \alpha

where

\tau is the net torque applied to the object

I is the moment of inertia

\alpha is the angular acceleration

In this problem we have:

\tau = -12.5 Nm (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

I=50.0 kg m^2 is the moment of inertia

Solving for \alpha, we find the angular acceleration:

\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2

3 0
3 years ago
An object is dropped from an altitude of one Earth radiusabove Earth's surface. If M is the mass of Earth and R is itsradius the
Montano1993 [528]

Answer:

The speed of the object just before it hits Earth is  \sqrt{\dfrac{GM}{R}}

(A) is correct option.

Explanation:

Given that,

M = mass of earth

R = radius of earth

The potential energy at height above the surface of the earth

P.E=-\dfrac{GmM}{R+h}

The kinetic energy at height above the surface of the earth

K.E = 0

The total energy at height above the surface of the earth

E = K.E+P.E

E = -\dfrac{GmM}{R+h}....(I)

The total energy at the surface of the earth

E'=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}....(II)

We need to calculate the speed of the object  just before it hits Earth

From equation (I) and (II)

-\dfrac{GmM}{R+h}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}

Here, h = R

-\dfrac{GmM}{2R}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}

v= \sqrt{\dfrac{GM}{R}}

Hence, The speed of the object just before it hits Earth is  \sqrt{\dfrac{GM}{R}}.

4 0
3 years ago
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