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2 years ago

object is placed 15cm in front of a plane mirror. If the mirror is moved further 5cm away from the object and the image is

Physics

2 answers:

4vir4ik [10]2 years ago

6 0

The object will be 20cm away because it was placed a further 5cm away from the mirror and 15 + 5 equals 20cm

solong [7]2 years ago

4 0

Answer:

The image will most likely be 20cm in front the mirror since the mirror was placed further 5cm.

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A car's gas tank contains 58.7 kg

Bogdan [553]

Answer:

721 kg/m^3

Explanation:

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2 years ago

A water bath in a physical chemistry lab is 1.55 m long, 0.710 m wide, and 0.570 m deep (high). If it is filled to within 3.55 i

Lesechka [4]

Answer:

528 liter.

Explanation:

Volume of the tank(cuboid) = l*b*h

But volume of the water = l*b*h

Where

l= length of the tank

b = width of the tank

h = the length from the bottom of the tank,

3.55 in to m,

0.09017m

Length of the water in the tank = 0.570 - 0.09017

= 0.47983 m.

Volume = 0.47983*0.710*1.55

= 0.528 m3.

1 m3 = 1000 liter.

0.528 m3 = 0.528*1000

= 528 liter

7 0

3 years ago

You observe lighting striking and then hear to sound 8 seconds later. The speed of sound in air is 340 m/s. How

Leto [7]

Answer:

2.72 Kilometers

Explanation:

8 × 340 m/s = 2720 m = 2.72 Kilometers

7 0

2 years ago

A children's roller coaster is released from the top of a track. If its maximum speed at ground level is 3 m/s, find the height

san4es73 [151]

Answer:

h = 0.46 m

Explanation:

According to the law of conservation of energy:

Potential Energy Lost by Roller Coaster = Kinetic Energy Gained by Roller Coaster

$mgh = \frac{1}{2}mv^2\\\\2gh = v^2\\\\h = \frac{v^2}{2g}$

where,

h = height = ?

v = speed at bottom = 3 m/s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$h = \frac{(3\ m/s)^2}{(2)(9.81\ m/s^2)}$

4 0

2 years ago

A 1.2 kg ball drops vertically onto a floor from a height of 32 m, and rebounds with an initial speed of 10 m/s.

iren [92.7K]

Explanation:

Given that,

Mass of the ball, m = 1.2 kg

Initial speed of the ball, u = 10 m/s

Height of the floor from ground, h = 32 m

(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 32}$

v = -25.04 m/s (negative as it rebounds)

The impulse acting on the ball is equal to the change in momentum. It can be calculated as :

$J=m(v-u)$

$J=1.2\times (-25.04-10)$

J = -42.048 kg-m/s

(b) Time of contact, t = 0.02 s

Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :

$J=\dfrac{F}{t}$

$F=J\times t$

$F=42.048\times 0.02$

F = 0.8409 N

Hence, this is the required solution.

5 0

2 years ago

object is placed 15cm in front of a plane mirror. If the mirror is moved further 5cm away from the object and the image is​

object is placed 15cm in front of a plane mirror. If the mirror is moved further 5cm away from the object and the image is