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2 years ago

1) Is air matter? Why or why not?

Physics

2 answers:

Zina [86]2 years ago

7 0

Answer:

air is matter

Explanation:

air is made up of molicules like liquids and solids making it matter

( matter rights matter)

Montano1993 [528]2 years ago

6 0

Answer: But, like solids and liquids, air is matter. It has weight (more than we might imagine), it takes up space, and it is composed of particles too small and too spread apart to see. Air, a mixture of gases, shares properties with water vapor, the gaseous form of water that is part of air.

Explanation:

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A flywheel of J = 50 kg-m2 initially standing still is subjected to a constant torque. If the angular velocity reaches 20 Hz in

Karolina [17]

Answer:

$\tau = 1256.5\ N.m$

Explanation:

given,

J = 50 kg-m²

frequency, f = 20 Hz

time ,t = 5 s

we know,

angular velocity = 2 π f

ω = 2 π x 20

ω = 125.66 rad/s

now, angular acceleration calculation

$\alpha = \dfrac{\omega_f-\omega_i}{t}$

$\alpha = \dfrac{125.66-0}{5}$

α = 25.13 rad/s²

Torque given to the flywheel.

$\tau = I \alpha$

$\tau = 50\times 25.13$

$\tau = 1256.5\ N.m$

Torque of the given flywheel is equal to $\tau = 1256.5\ N.m$

7 0

3 years ago

A spring stretches 0.018 m when a 2.8-kg object is suspendedfrom its end. How much mass should be attached to this spring sothat

agasfer [191]

Answer:

m = 4.29 kg

Explanation:

Given that,

Mass of the object, m = 2.8 kg

Stretching in the spring, x = 0.018 m

Frequency of vibration, f = 3 Hz

Let m is the mass of the object that is attached to the spring. When it is attached the gravitational force is balanced by the force on spring. It is given by :

$mg=kx$

$k=\dfrac{mg}{x}$

$k=\dfrac{2.8\times 9.8}{0.018}$

k = 1524.44 N/m

Since, $\omega=\sqrt{\dfrac{k}{m}}$

$m=\dfrac{k}{4\pi^2 f^2}$

$m=\dfrac{1524.44}{4\pi^2 \times 3^2}$

m = 4.29 kg

So, the mass that is attached to this spring is 4.29 kg. Hence, this is the required solution.

4 0

3 years ago

A girl with a mass of 27 kg is playing on a swing. There are three main forces

N76 [4]

The tension in the swing's chain at the bottom of the swing is 178.35 N.

The given parameters:

Mass of the girl, m = 27 kg
Speed of the girl, v = 3 m/s
Radius of the circle, r = 4 m

The tension in the swing's chain at the bottom of the swing is calculated as follows;

$T = mg + ma_c\\\\ T= mg + \frac{mv^2}{r} \\\\T = (12 \times 9.8) + (\frac{27 \times 3^2}{4} )\\\\T = 117.6 \ N \ + \ 60.75 \ N\\\\T = 178.35 \ N$

Thus, the tension in the swing's chain at the bottom of the swing is 178.35 N.

Learn more about tension in vertical circle here: brainly.com/question/19904705

3 0

2 years ago

330 grams of boiling water (temperature 100°C, specific heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass i

Alexus [3.1K]

Answer:

T = 74°C

Explanation:

Given Mw = mass of water = 330g, Ma = mass of aluminium = 840g

Cw = 4.2gJ/g°C = specific heat capacity of water and Ca = 0.9J/g°C = specific heat capacity of aluminium

Initial temperature of water = 100°C.

Initial temperature of aluminium = 29°C

When the boiling water is poured into the aluminum pan, heat is exchanged and after a short time the water and aluminum pan both come to thermal equilibrium at a common temperature T.

Heat lost by water equal to the heat gained by aluminium pan.

Mw × Cw×(100 –T) = Ma × Ca × (T–29)

330×4.2×(100– T) = 890×0.9×(T–29)

1386(100 – T) = 801(T –29)

1386/801(100 – T) = T – 29

1.73(100 – T) = T – 29

173 –1.73T = T –29

173+29 = T + 1.73T

202 = 2.73T

T = 202/2.73

T = 74°C

6 0

3 years ago

If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. Wha

Snowcat [4.5K]

Answer:

176.4 m

Explanation:

U = 0, t = 6s, g = 9.8 m/s^2

Use second equation of motion

H = ut + 1/2 gt^2

H = 0 + 0.5 × 9.8 × 6 × 6

H = 176.4 m

It is the displacement from the point of dropping of object.

3 0

3 years ago