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Elanso [62]
3 years ago
12

A car moving at 10.0 m/s encounters a bump that has a circular cross-section with a radius of 30.0 m. What is the normal force e

xerted by the seat of the car on a 60.0-kg passenger when the car is at the top of the bump?
A) 200 N
B) 389 N
C) 789 N
D) 589 N
Physics
2 answers:
Andrej [43]3 years ago
5 0

Answer:

option B

Explanation:

given.

car is moving at = 10 m/s

bump of radius = 30 m

normal force = ?

mass of the passenger = 60 kg

m × g = 60 × 9.81 N

normal force acting

                 =mg - \dfrac{mv^2}{R}

                 =60\times 9.81 - \dfrac{60\times 10^2}{30}

                 =388.6 N

the normal force acting is equal to 389 N

hence, the correct answer is option B

Andrew [12]3 years ago
4 0

Answer:

force at the top will be 388 N which is nearly equal to 389

option (b) is correct

Explanation:

We have given velocity of the moving car v = 10 m/sec

Radius of the circular section r = 30 m

Mass of the passenger m = 60 kg

Acceleration due to gravity g=9.8m/sec^2

At the top normal force is given by noraml\ force=mg-\frac{mv^2}{r}

So force at top will be F=60\times 9.8-\frac{60\times 10^2}{30}=388N

So force at the top will be 388 N which is nearly equal to 389

So option (b) is correct

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When this current is closed which way does the current flow
Anastaziya [24]
Well, Godess, that's not a simple question, and it doesn't have
a simple answer.

When the switch is closed . . .

"Conventional current" flows out of the ' + ' of the battery, through R₁ ,
then through R₂ , then through R₃ .  It piles up on the right-hand side of
the capacitor (C).  It repels the ' + ' charges on the left side of 'C', and
those flow into the ' - ' side of the battery.  So the flow of current through
this series circuit is completely clockwise, around toward the right. 

That's the way the first experimenters pictured it, that's the way we still
handle it on paper, and that's the way our ammeters display it.

BUT . . .

About 100 years after we thought that we completely understand electricity,
we discovered that the little tiny things that really move through a wire, and
really carry the electric charge, are the electrons, and they carry NEGATIVE
charge.  This turned our whole picture upside down.

But we never changed the picture !  We still do all of our work in terms of
'conventional current'.  But the PHYSICAL current ... the actual motion of
charge in the wire ... is all exactly the other way around.

In your drawing ... When the switch is closed, electrons flow out of the 
' - ' terminal on the bottom of the battery, and pile up on the left plate of
the 'C'.  They repel electrons off of the right-side of 'C', and those then
flow through R₃ , then through R₂ , then through R₁ , and finally into the
' + ' terminal on top of the battery.

Those are the directions of 'conventional' current and 'physical' current
in all circuits.

In the circuit of YOUR picture that you attached, there's more to the story:

Battery current can't flow through a capacitor.  Current flows only until
charges are piled up on the two sides of 'C' facing each other, and then
it stops.

Wait a few seconds after you close the switch in the picture, and there is
no longer any current in the loop.

To be very specific and technical about it . . .

-- The instant you close the switch, the current is

       (battery voltage) / (R₁ + R₂ + R₃)        amperes

but it immediately starts to decrease.

--  Every  (C)/((R₁ + R₂ + R₃)  seconds after that, the current is

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less than it was that same amount of time ago.

Now, are you glad you asked ?
4 0
4 years ago
Find the velocity (in m/s) of a proton that has a momentum of 4.96 X 10^-19 kg.m/s.
kramer

Answer:

The velocity of the proton is 2.965*10^{8} \frac{m}{s}

Explanation:

The momentum of a particle is defined as the product of its mass by its velocity and we can calculate it using the following formula:

p=m*v Equation (1)

Where:

p: Is the momentum in kg*m/s

m: Is mass of the particle in kg

v: Is the velocity of the particle in m/s

Data known:

m= 1,6726 × 10^–27 kg : mass of the proton

p= 4.96 X 10^-19 kg.m/s.

We replace this data in the Equation (1):

4,96*10^{-19} =1.6726*10^{-27} *vv=\frac{4.9*10^{-19} }{1.6726*10^{-27} }

v=(\frac{4.96}{1.6726} )*(10^{27-19} )

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3 0
3 years ago
Superman takes 8 seconds to stop a runway train over a distance of 58 m using a power of
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Answer:

Workdone = 600 Kilojoules

Explanation:

Given the following data:

Time = 8 seconds

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Distance = 58 m

To find the work done;

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

Power = \frac {Energy}{time}

Thus, work done is given by the formula;

Workdone = power * time

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Workdone = 600,000 = 600 KJ

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