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2 years ago

How could you make a feather and a rubber ball hit the ground at the same time?

Physics

1 answer:

raketka [301]2 years ago

3 0

Answer:

If you take out air they'll fall at the same time

Explanation:

for example a closed umbrella will fall faster than an open one because the air is holding the opened umbrella back

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Name 3 constellations that are on the ecliptic and visible at this time

Hoochie [10]

Answer:

also right now we are in ZODIAC SEASON OF Libra WE JUST PASSED VIRGO AND WE ARE HEADING INTO the constellations of fall are Aquarius, Grus, Lacerta, Octans, Pegasus and Piscis Austrinus. Lacerta and Pegasus are located in the northern sky

Explanation:

The five northern constellations visible from most locations north of the equator throughout the year are Cassiopeia, Cepheus, Draco, Ursa Major, and Ursa Minor.

The three southern circumpolar constellations visible from most locations in the southern hemisphere are Carina, Centaurus, and Crux.

The ecliptic currently passes through the following constellations:

Pisces.

Aries.

Taurus.

Gemini.

Cancer.

Leo.

Virgo.

Libra. hope it helps

6 0

4 years ago

If you open the valve on this bicycle tire what would happen to the air inside of the tire why do you think that would happen

Flauer [41]

The oxygen will flow out of the tire,
which causes the tire to become flat.

8 0

3 years ago

A battery with an emf of 12.0 V shows a terminal voltage of 11.8 V when operating in a circuit with two lightbulbs rated at 3.0

Blababa [14]

Answer:

$R = 24 ohm$

Explanation:

As we know that terminal voltage and EMF of cell is given as

$V = EMF - ir$

$ir = 12 - 11.8$

$ir = 0.2 Volts$

resistance of two bulbs is given as

$R = \frac{V^2}{P}$

$R = \frac{12^2}{3}$

$R = 48 ohm$

now these two bulbs are connected in parallel

so equivalent resistance is given as

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{1}{R} = \frac{1}{48} + \frac{1}{48}$

so we have

$R = 24 ohm$

3 0

3 years ago

A hamster of mass 0.139 kg gets onto his 20.8−cm-diameter exercise wheel and runs along inside the wheel for 0.823 s until its f

Anna11 [10]

Explanation:

Given that,

Mass of the hamster, m = 0.139 kg

Diameter of the wheel, d = 20.8 cm

Radius, r = 10.4 m

Frequency of the wheel, f = 1 Hz

Time, t = 0.823 s

(a) There exist a relationship between the tangential and angular acceleration of the wheel. It is given by :

$a=\alpha \times r$

Since, $\alpha =\dfrac{\omega}{t}$

$a=\dfrac{\omega}{t} \times r$

$a=\dfrac{2\pi f}{t} \times r$

$a=\dfrac{2\pi \times 1}{ 0.823} \times 10.4\times 10^{-2}$

$a=0.793\ m/s^2$

(b) In radial direction, applying the second law of motion as :

$N-mg=ma$

a is the radial acceleration, $a=\dfrac{v^2}{r}$

$N=mg+ma$

$N=mg+m(\dfrac{v^2}{r})$

$N=mg+m(\dfrac{(r\omega)^2}{r})$

$N=mg+m\omega^2 r$

$N=m(g+\omega^2 r)$

$N=m(g+(2\pi f)^2 r)$

$N=0.139\times (9.8+(2\pi 1)^2\times 10.4\times 10^{-2})$

$N=1.93\ N$

Hence, this is the required solution.

8 0

4 years ago

A 0.18-kg turntable of radius 0.32 m spins about a vertical axis through its center. A constant rotational acceleration causes t

borishaifa [10]

Answer:

Angular acceleration will be $18.84rad/sec^2$

Explanation:

We have given that mass m = 0.18 kg

Radius r = 0.32 m

Initial angular velocity $\omega _i=0rev/sec$

And final angular velocity $\omega _f=24rev/sec$

Time is given as t = 8 sec

From equation of motion

We know that $\omega _f=\omega _i+\alpha t$

$24=0+\alpha \times 8$

$\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2$

So angular acceleration will be $18.84rad/sec^2$

4 0

4 years ago