Please help, question is timed...

Answers

Answer 1

Answer:

Answer:

Explanation:

50+3*5

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The acceleration of an object increases by the force applied to it and decreases based on it's mass.

Answers

Explanation:

Force is defined as mass times acceleration.

Mathematically, F = m × a

where F = force

m = mass

a = acceleration

This shows that force is directly proportional to acceleration. So, when greater force is applied then acceleration will also increase.

Whereas acceleration is inversely proportional to mass. So, when there is increase in mass then there will be decrease in acceleration.

When this current is closed which way does the current flow

Answers

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

e⁻¹ = about 36.8 %

less than it was that same amount of time ago.

Now, are you glad you asked ?

A 250 gram ball at the end of a string is revolving uniformly in a circle of radius 0.75 meters.The ball makes 2.0 revolutions per second. What is the centripetal acceleration?

Answers

The centripetal acceleration of the ball would be 88.44 m/ $s^2$ .

Centripetal acceleration

The centripetal acceleration (ac) of an object moving in a circle at a constant speed is given by the formula:

ac = (v^2) / r

where v is the speed of the object and r is the radius of the circle.

In this case, the ball is revolving uniformly in a circle of radius 0.75 meters, and it makes 2.0 revolutions per second. To find the speed of the ball (v), we need to convert the number of revolutions per second to the angular velocity (ω) in radians per second:

ω = 2π x (number of revolutions per second)

ω = 2π x 2.0 = 4π radians per second

The speed of the ball (v) is then given by:

v = ω x r
v = (4π rad/s) x 0.75 m = 3π m/s

Now we can calculate the centripetal acceleration (ac) of the ball:

ac = (v^2) / r
ac = [(3π m/s)^2] / 0.75 m
ac = 9π^2 m/s^2 ≈ 88.44 m/s^2

Therefore, the centripetal acceleration of the ball is approximately 88.44 m/s^2.

More on centripetal acceleration can be found here: brainly.com/question/1052267

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Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at a rate of 20,000 Btu/ hr., what is the temperature of the water leaving the radiator?

Answers

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

A speedy rabbit is hopping to the right with a velocity of 4.0 \,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction when it sees a carrot in the distance. The rabbit speeds up to its maximum velocity of 13 \,\dfrac{\text m}{\text s}13 s m 13, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction with a constant acceleration of 2.0 \,\dfrac{\text m}{\text s^2}2.0 s 2 m 2, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction rightward.

Answers

Answer: 38.25 m

Explanation:

In this situation we need to find the distance $d$ between the rabbit and the carrot, and we can use the following equation, since the rabbit's acceleration is constant: