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3 years ago

100 POINTS! PLEASE HELP!

Physics

2 answers:

bagirrra123 [75]3 years ago

8 0

Answer:

By the information given, I am not sure how the event has played out. However, using common knowledge, I can give a scenario of how it may play out.

When removing the first block (the foundation), gravity will pull the other blocks down, as the strength of gravity exceeds that of the strength of air holding it up. This means that all the blocks on top would fall.

By definition of Newton's 1st - 3rd laws, the blocks would not move unless a force is given to it (i.e., removing the bottom block, which causes gravity). Gravity would pull the other blocks downwards. This would lead to the falling of the stack of blocks.

vaieri [72.5K]3 years ago

8 0

Answer:

Explanation:

A stack of blocks sits in equilibrium. That means, by Newton's 1st law, all blocks will remain stationary. The weight of each block is balanced by the reaction force from the block underneath it. The bottom block experiences a reaction force equals to the total weight of all blocks from the ground.

When the bottom block is removed, there will not be any reaction force from the ground. That causes a disturbance in the equilibrium. Without the balancing reaction, the blocks will have the gravity force from their weights pulling them down. By Newton 2nd law, the unbalance force will cause the blocks to accelerate towards the ground until they all crash-land.

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Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and

dexar [7]

Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
closest answer is 1.99
× 10^30

(it may vary
a little with rounding – the difference is less than 1%)

8 0

3 years ago

Read 2 more answers

At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re

densk [106]

Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a) As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ $-( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}$

where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

$v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s$

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

$-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}$

where e= elementary charge= 1.6*10⁻¹⁹ C, VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

$v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s$

⇒ ve = 1.08*10⁶ m/s

4 0

3 years ago

Four compasses are placed on a table top

motikmotik

Answer:

located diagonal

Explanation:

check ur text book bro

7 0

2 years ago

Forces normal to a particle's displacement do no work.

snow_tiger [21]

Answer:

Explanation:

The work done is defined as the product of force applied in the direction of displacement and the displacement.

W = F x d x Cosθ

where, F is the force applied, d be the displacement and θ be the angle between the displacement and force.

For the normal forces, the angle between the displacement and the force applied is 90 degree, and the value of Cos 90 is zero, so the work done is zero.

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3 years ago

The charges that are free to move in a metallic conducting wire and that are responsible for the flow of electric current are- a

SVEN [57.7K]

Answer:

D) The negatively charged electrons

Electricity passes through metallic conductors as a flow of negatively charged electrons. The electrons are free to move from one atom to another. We call them a sea of delocalised electrons. Current was originally defined as the flow of charges from positive to negative. Please give me the brainliest answer?

:) Hoped this helped!!! Have a good day!!! <3

3 0

2 years ago