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3 years ago

If a 100N weight sits in an express elevator what will be the weight when the elevator is traveling at a constant speed

1 answer:

4 0

This is a "trick" question.

If the elevator is traveling at constant speed, it means it is at rest. This means anything inside the elevator traveling at constant speed, weights the same as in an elevator not moving -also at rest-.

So the 100N weight's weight doesn't change in an elevator traveling at constant speed.

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An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?

Scorpion4ik [409]

Answer:

80 ft/s

Explanation:

Use III equation of motion

V^2 = U^2 + 2g h

Here, U = 0, g = 32 ft/s^2, h = 100 ft

V^2 = 0 + 2 × 32 ×100

V^2 = 6400

V = 80 ft/s

8 0

2 years ago

What The BEST GAME?????/ a: Roblox b; Mincraft c; Fortnite

Maksim231197 [3]

Answer:

A,C

Explanation:

8 0

2 years ago

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The images show two different types of Galapagos tortoises that scientists believe descended from the same species. The first ty

vichka [17]

Answer:

C. turtles with genes for long necks had a better chance of surviving to reach reproductive age.

Explanation:

The turtles that had long necks were more fit to the environmnet in which they were lovated and were able to grow larger and have more reproductive time because of their ability to feed on grass and small shrubs, this helped them always haev food available, and made them the dominant gene eventually.

3 0

3 years ago

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A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.

lana [24]

In this problem we have the electric field intensity E:

E = 6.5 × $10^4$ newtons/coulomb

We have the magnitude of the load:

q = 6.4 × $10 ^{-19}$ coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × $10^{-2}$ meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × $10^{-19}$ )(6.5 × $10^4$ )(1.2 × $10^{-2}$ )

PE = 5.0 x $10^{-16}$ joules

None of the options shown is correct.

6 0

2 years ago

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

liq [111]

Answer:

The speed of water must be expelled at 6.06 m/s

Explanation:

Neglecting any drag effects of the surrounding water we can assume the linear momentum in this case is conserves, that is, the total initial momentum of the octopus and the water kept in it cavity should be equal to the total final linear momentum. That's known as conservation of momentum, mathematically expressed as:

$p_f=p_i$

with Pi the total initial momentum and Pf the final total momentum. The total momentum is the sum of the momentums of the individual objects, in our case the octopus and the mass of water that will be expelled:

$p_{of}+p_{wf}=p_{oi}+p_{wi}$

with Po the momentum of the octopus and Pw the momentum of expelled water. Linear momentum is defined as mass times velocity:

$m_o*v_{of}+m_w*v_{wf}=m_o*v_{oi}+m_w*v_{wi}$

Note that initially the octopus has the water in its cavity and both are at rest before it sees the predator so $v_{oi}=v_{wi} = 0\frac{m}{s}$ :

$m_o*v_{of}+m_w*v_{wf}=0$

We should find the final velocity of water if the final velocity of the octopus is 2.70 m/s, solving for $v_{wf}$ :

$v_{wf}=-\frac{m_o*v_{of}}{m_w}=-\frac{(6.00-1.85)*(2.70)}{1.85}$

$v_{wf}=-6.06\frac{m}{s}$

The minus sign indicates the velocity of the water is opposite the velocity of the octopus.

3 0

2 years ago