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2 years ago

WHEN THE FORCE APPLIED ON AN ELASTIC OBJECT IS EQUAL AND OPPOSITE,THEN THIS FORCE MAY

Physics

1 answer:

jok3333 [9.3K]2 years ago

4 0

Answer:

Elastic force

Explanation:

Elastic force occur when deformed or elastic object try to return to its original or natural length. When an elastic material is stretched, it resist it change in shape. The force act to return the elastic material to its original shape, it's then exert a counter force in opposit e direction.

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I.Name two commonly used thermometric liquids.

PtichkaEL [24]

Answer:

mercury and alcohol

ii) used to test temperatures

6 0

2 years ago

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A 50 Q resistor in a circuit has a current flowing through it of 2.0 A. What is

Alex17521 [72]

Hello!

We can use the following equation for calculating power dissipated by a resistor:
$P = i^2R$

P = Power (? W)
i = Current through resistor (2.0 A)
R = Resistance of resistor (50Ω)

Plug in the known values and solve.

$P = (2.0^2)(50) = \boxed{\text{ B. }200 W}$

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1 year ago

7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee

wlad13 [49]

Answer:

$N_{top}=9.8N\\N_{bottom}=33.4N$

b) $v_{min}=4.4m/s$

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is $a_{cp}=\frac{v^2}{R}$ . At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

$-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg$

Which means:

$N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N$

The limit for falling off would be $N_{top}=0$ , so the minimum speed would be:

$0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s$

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2 years ago

What was your electric potential relative to a metal pipe if a spark jumped 1.1 cm through dry air from your finger to the pipe?

romanna [79]

The electric field is given by volts/distance: $E= \frac{V}{d}$ . The breakdown voltage of dry air is about 3x10^6V/m. So solving for V we get
$V=Ed$
or $V=(3e6V/m)(0.011m)=33,000V$

6 0

3 years ago

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which has a greater kinetic energy, a bowling ball that has a mass of 5kg travelling at 6m/s, or a ship that has a mass of 12000

AVprozaik [17]

Answer:bowling ball has greater kinetic energy

Explanation:

Kinetic energy of bowling ball:

mass=m=5kg

Velocity=v=6m/s

Kinetic energy =ke

Ke=0.5 x m x v x v

Ke=0.5 x 5 x 6 x 6

Ke=90J

Kinetic energy of ship:

mass=m=120000kg

velocity=v=0.02m/s

Ke=0.5 x m x v x v

Ke=0.5 x 120000 x 0.02 x 0.02

Ke=24J

6 0

2 years ago