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3 years ago

When we place an opaque object between a source of light and screen we will observe

Physics

1 answer:

Alex_Xolod [135]3 years ago

3 0

No light reaches within the geometrical shadow of the obstacle at the screen.<span>
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May somebody help me in this.

aliina [53]

Answer:

answer c. come easy dear.

7 0

3 years ago

One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched

mash [69]

Answer:

2.63 cm

Explanation:

Hooke's law gives that the force F is equal to cy where c is spring constant and x is extension

Making c the subject of the formula then

$c=\frac {F}{y}$

Since F is gm but taking the given mass to be F

$c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930$

By substitution now considering F to be 3.3 kg

$y=\frac {3.3 kg}{1.2558139534883720930232558139534883720930}=2.6277777777777 cm\approx 2.63 cm$

8 0

4 years ago

The weld current between pulses is know as ______

Brut [27]

Answer:

the pulse arc system

Explanation:

6 0

3 years ago

A department store sells an ""astronomical telescope"" with an objective lens of 30 cm focal length and an eyepiece lens of foca

Yuliya22 [10]

Answer:

The magnifying power of this telescope is (-60).

Explanation:

Given that,

The focal length of the objective lens of an astronomical telescope, $f_o=30\ cm$

The focal length of the eyepiece lens of an astronomical telescope, $f_e=5\ mm=0.5\ cm$

To find,

The magnifying power of this telescope.

Solution,

The ratio of focal length of the objective lens to the focal length of the eyepiece lens is called magnifying of the lens. It is given by :

$m=\dfrac{-f_o}{f_e}$

$m=\dfrac{-30}{0.5}$

m = -60

So, the magnifying power of this telescope is 60. Therefore, this is the required solution.

7 0

3 years ago

A holiday ornament in the shape of a hollow sphere with mass 0.015 kg and radius 0.055 m is hung from a tree limb by a small loo

s344n2d4d5 [400]

Answer: 0.61 s

Explanation:

Given

Mass of object, m = 0.015 kg

Radius of object, r = 0.055 m

Acceleration of object, g = 9.8 m/s²

In a pendulum,

T = 2π * √[I /(mgd)]

The moment of Inertia, I of a hollow sphere is given by

I(sphere) = 2/3MR² + MR²

I(sphere) = 5/3MR²

Also, d = R

Substituting these into the first equation, we have

T = 2π * √[(5/3MR²) / (mgr)]

T = 2π * √[(5/3r) / (g)]

T = 2 * 3.142 * √(5/3 * 0.055) / (9.8)]

T = 6.284 * √(0.092/9.8)

T = 6.284 * √0.00939

T = 6.284 * 0.097

T = 0.6095 s

To 2 significant figures,

The period is 0.61 s

7 0

3 years ago