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natka813 [3]
3 years ago
13

Can someone help me plzit's a glass bell jar and to vacuum pump

Physics
1 answer:
Nata [24]3 years ago
7 0
A) the pump removes air from the enclosure. foam absorbs vibrations from the clock
b) result is that as the air is taken out by pump, the sound of alarm reduces and eventually becomes zero. so, you cant hear the alarm ringing.
c) conclusion is that sound cant travel in vaccum and it requires a medium to travel.
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Suggest two uses of pulleys.
Marizza181 [45]

Answer: Elevators use pulleys to function.

A cargo lift system that allows for items to be hoisted to higher floors is a pulley system.

Explanation:

8 0
2 years ago
On Earth, a spring stretches by 5.0 cm when a mass of 3.0 kg is suspended from one end.
Neko [114]

Answer:

Mass = 18.0 kg

Explanation:

From Hooke's law,

F = ke

where: F is the force, k is the spring constant and e is the extension.

But, F = mg

So that,

mg = ke

On the Earth, let the gravitational force be 10 m/s^{2}.

3.0 x 10 = k x 5.0

30 = 5k

⇒ k = \frac{30}{5} ................ 1

On the Moon, the gravitational force is \frac{1}{6} of that on the Earth.

m x \frac{10}{6} = k x 5.0

\frac{10m}{6} = 5k

⇒ k = \frac{10m}{30} ............. 2

Equating 1 and 2, we have;

\frac{30}{5}  = \frac{10m}{30}

m = \frac{900}{50}

    = 18.0

m = 18.0 kg

The mass required to produce the same extension on the Moon is 18 kg.

8 0
3 years ago
Read 2 more answers
Tell the value of the underlined digit 843,208,732,833 eight is underlined
devlian [24]
The value of the underlined 8 is, hundred billion's. Hope this helped!
4 0
3 years ago
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An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

6 0
3 years ago
En el diseño de una sesión de aprendizaje, la profesora Patricia, tiene como propósito de aprendizaje: Aplica sus habilidades mo
Leviafan [203]

Answer:

im in hardvard dont cheat get amzing grades if you want to go here

Explanation:

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3 years ago
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