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3 years ago

Explain why the open umbrella hit the ground after the closed umbrella. *

Physics

1 answer:

balandron [24]3 years ago

7 0

Answer:

An open umbrella would not hit the ground in front of the closed one.

Explanation:

The closed umbrella would impact first because it did not have as much air pressure. The open one has air that presses against it and makes it reach where the open umbrella floats but falls slowly

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A tiger leaps with a horizontal speed of 4.5 m/s from a boulder and lands 15 meters away.What is the vertical velocity with whic

Gelneren [198K]

Answer:

v_oy = 16.33 m/s

Explanation:

To find the vertical velocity of the tiger, you use the information about the horizontal velocity and maximum horizontal distance traveled.

You use the following formula for the range of the trajectory:

$x_{max}=\frac{2v_{ox}v_{oy}}{g}$ ( 1 )

v_ox: horizontal initial velocity = 4.5m/s

v_oy: vertical initial velocity = ?

g: gravitational acceleration = 9.8m/s^2

x_max: range of the trajectory = 15 m

You do v_oy the subject of the formula ( 1 ) and you replace the values of the other parameters in order to calculate v_oy:

$v_{oy}=\frac{gx_{max}}{2v_{ox}}=\frac{(9.8m/s^2)(15m)}{2(4.5m/s)}\\\\v_{oy}=16.33\frac{m}{s}$

hence, the initial vertical velocity of the tiger is 16.33m/s

8 0

3 years ago

A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn

EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, $C=5.00 \mu F$ , the charged stored on the capacitor at the end of the process is

$Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C$

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

$V=\frac{Q}{C}$

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

$C=\frac{\epsilon_0 \epsilon_r A}{d}$

where:

$\epsilon_0$ is the permittivity of free space

$\epsilon_r$ is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: $C'=\frac{C}{2}$ . The new voltage across the plate is given by

$V'=\frac{Q}{C'}$

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V$

So, the new voltage is

$V'=2 (12.0 V)=24.0 V$

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

$A=\pi r^2$

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

$A'=\pi (2r)^2 = 4 \pi r^2 = 4A$

While the separation between the plate was unchanged (d); so, the new capacitance will be

$C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C$

So, the capacitance has increased by a factor 4; therefore, the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}$

which means

$V'=\frac{12.0 V}{4}=3.0 V$

3 0

3 years ago

A college student likes to take her books to class by placing them in a box and pulling the box around behind her. She pulls on

just olya [345]

Answer:

It would result an a negatice answer.

Explanation:

The accelarion should be pulled as a kite not a box :) columbus said that musical stuff no just no

Hope this helps!!

- Katty queen

8 0

3 years ago

PHYSICS CIRCUIT QUESTION PLEASE HELP!! 20 Points!

dimulka [17.4K]

This really calls for a blackboard and a hunk of chalk, but
I'm going to try and do without.

If you want to understand what's going on, then PLEASE
keep drawing visible as you go through this answer, either
on the paper or else on a separate screen.

The energy dissipated by the circuit is the energy delivered by
the battery. We'd know what that is if we knew I₁ . Everything that
flows in this circuit has to go through R₁ , so let's find I₁ first.

-- R₃ and R₄ in series make 6Ω.
-- That 6Ω in parallel with R₂ makes 3Ω.
-- That 3Ω in series with R₁ makes 10Ω across the battery.
-- I₁ is 10volts/10Ω = 1 Ampere.

-- R1: 1 ampere through 7Ω ... V₁ = I₁ · R₁ = 7 volts .

-- The battery is 10 volts.
    7 of the 10 appear across R₁ .
   So the other 3 volts appear across all the business at the bottom.

-- R₂: 3 volts across it = V₂.
   Current through it is I₂ = V₂/R₂ = 3volts/6Ω = 1/2 Amp.

-- R3 + R4: 6Ω in the series combination
   3 volts across it
   Current through it is I = V₂/R = 3volts/6Ω = 1/2 Ampere

-- Remember that the current is the same at every point in
a series circuit. I₃ and I₄ must be the same 1/2 Ampere,
because there's no place in the branch where electrons can
be temporarily stored, no place for them to leak out, and no
supply of additional electrons.

-- R₃: 1/2 Ampere through it = I₃ .
   1/2 Ampere through 2Ω ... V₃ = I₃ · R₃ = 1 volt

-- R₄: 1/2 Ampere through it = I₄
   1/2 Ampere through 4Ω ... V₄ = I₄ · R₄ = 2 volts

Notice that I₂ is 1/2 Amp, and (I₃ , I₄) is also 1/2 Amp.
So the sum of currents through the two horizontal branches is 1 Amp,
which exactly matches I₁ coming down the side, just as it should.
That means that at the left side, at the point where R₁, R₂, and R₃ all
meet, the amount of current flowing into that point is the same as the
amount flowing out ... electrons are not piling up there.

Concerning energy, we could go through and calculate the energy
dissipated by each resistor and then addum up. But why bother ?
The energy dissipated by the resistors has to come from the battery,
so we only need to calculate how much the battery is supplying, and
we'll have it.

The power supplied by the battery = (voltage) · (current)

   = (10 volts) · (1 Amp) = 10 watts .

"Watt" means "joule per second".
The resistors are dissipating 10 joules per second,
and the joules are coming from the battery.

   (30 minutes) · (60 sec/minute) = 1,800 seconds

   (10 joules/second) · (1,800 seconds) = 18,000 joules in 30 min

The power (joules per second) dissipated by each individual resistor is

   P = V² / R
   or
   P = I² · R ,

whichever one you prefer. They're both true.

If you go through the 4 resistors, calculate each one, and addum up, you'll
come out with the same 10 watts / 18,000 joules total.

They're not asking for that. But if you did it and you actually got the same
numbers as the battery is supplying, that would be a really nice confirmation
that all of your voltages and currents are correct.

7 0

3 years ago

Which scenarios are examples of physical change?

Oxana [17]

Answer:

For example: Freezing, boiling, are physical

Explanation:

3 0

4 years ago

Read 2 more answers