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2 years ago

Why can concave mirror is used in cosmetic mirror

Physics

1 answer:

tatuchka [14]2 years ago

4 0

Answer:

"When face is placed between the concave mirror and its focus, it produces a magnified image. This enlarged image of face is helpful in makeup as even pores of skin are clearly visible."

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A world-class sprinter can burst out of the blocks to essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race.

wolverine [178]

Given:
u = 0, initial speed (sprinter starts from rest)
v = 11.5 m/s, final speed
s = 15 m, distance traveled to attain final speed.

Let
a = average acceleration,
t = time taken to attain final speed.

Then
v² = u² + 2as
or
(11.5 m/s)² = 2*(a m/s²)*(15 m)
a = 11.5²/(2*15) = 4.408 m/s²

Also
v = u +a t
or
(11.5 m/s) = (4.408 m/s²)*(t s)
t = 11.5/4.408 = 2.609 s

Answer:
The average acceleration is 4.41 m/s² (nearest hundredth).
The time required is 2.61 s (nearest hundredth).

8 0

2 years ago

A box of paper is labeled 24lb paper. This means that 500 sheets (counted number) of paper size 17in x 22in weighs 24 pounds (Th

Tresset [83]

Answer:

The mass of a single paper is approximately 0.047 lb/paper which in SI Units is approximately 21.77 g/paper

Explanation:

The given information on the size and the weight of paper are;

The mass of a box of 500 sheets of paper = 24 lb

The number of sheets in the paper = 500 sheets

The dimensions of the paper = 17 in. × 22 in., which is equivalent to 43.18 cm × 55.88 cm

The mass of a single paper = The mass of the box of paper/(The number of sheets of paper present in the box)

The mass of a single paper = 24 lb/500 = 0.047 lb/paper

Given that 1 lb = 453.6 g, we have;

0.047 lb/paper = 0.047 lb/paper×453.6 g/(lb) = 21.77 g/paper

The mass of a single paper = 0.047 lb/paper = 21.77 g/paper.

6 0

2 years ago

A railroad car of mass 2.50 3 104 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railro

ipn [44]

Answer:2.5 m/s

37.5KJ

Explanation:

Let $u_1, u_2 , v_f$ be the initial velocity of rail road car ,coupled cars & Final velocity of system respectively.

$m=2.50\times 10^{4}$

Conserving momentum

$mu_1+3mu_2=4mv_f$

$4m+6m=4mv_f$

$v_f=2.5 m/s$

Therefore Final velocity of system is 2.5m/s

(b)Mechanical Energy lost =Initial Kinetic Energy -Final Kinetic Energy

$Initial Kinetic Energy=\frac{1}{2}m\left ( 4^2\right )+\frac{1}{2}m\left ( 2^2\right )=14m J$

$Final Kinetic Energy=\frac{1}{2}4m\left ( 2.5^2\right )=12.5m J$

$Mechanical Energy lost=14m-12.5m=3.75\times 10^4=37.5 KJ$

4 0

3 years ago

What is a combination of two or more simple machines?

Anna [14]

A compound

Explanation:

A compound machine is the combination of two or more simple machines.

An example is bicycle.

A simple machine is a device that is used to increase the magnitude of force.
It is a basic mechanical unit.
Examples are inclined planes, lever systems, wheel and axle.
A compound machine is a combination of these simple machines.

learn more:

lever brainly.com/question/5352966

#learnwithBrainly

7 0

2 years ago

A 10,000-kg railroad car travels down the track at 30 m/s and hits a second stationary railroad car of 20,000 kg. The cars becom

Ivanshal [37]

Answer:

10 m/s

Explanation:

The problem can be solved by using the law of conservation of momentum: the initial momentum has to be equal to the final momentum, so we can write the following

$p_i = p_f$