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lorasvet [3.4K]
3 years ago
11

the density of dry air at 20 degrees celsius is 1.20 g/L. What is the mass of air, in kilograms, of a room that measures 24.0m b

y 15.0 m by 4.0 m?
Physics
1 answer:
horsena [70]3 years ago
8 0
<h2>Mass of air in a room that measures 24.0 m by 15.0 m by 4.0 m is 1728 kg.</h2>

Explanation:

Density of air = 1.20 g/L = \frac{1.20\times 10^{-3}kg}{10^{-3}m^3}=1.2kg/m^3

Size of room is 24.0m by 15.0 m by 4.0 m

Volume of room = 24 x 15 x 4 = 1440 m³

We know the equation

            Mass = Volume x Density

            Mass = 1440 x 1.2

            Mass = 1728 kg  

Mass of air in a room that measures 24.0 m by 15.0 m by 4.0 m is 1728 kg.

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Which of the following objects is exerting a gravitational force on the floating tool? the Earth, the Moon, the astronaut, the S
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All of them are correct.
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2 years ago
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A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other mov
Brums [2.3K]

Answer: 330.88 J

Explanation:

Given

Linear velocity of the ball, v = 17.1 m/s

Distance from the joint, d = 0.47 m

Moment of inertia, I = 0.5 kgm²

The rotational kinetic energy, KE(rot) of an object is given by

KE(rot) = 1/2Iw²

Also, the angular velocity is given

w = v/r

Firstly, we calculate the angular velocity. Since it's needed in calculating the Kinetic Energy

w = v/r

w = 17.1 / 0.47

w = 36.38 rad/s

Now, substituting the value of w, with the already given value of I in the equation, we have

KE(rot) = 1/2Iw²

KE(rot) = 1/2 * 0.5 * 36.38²

KE(rot) = 0.25 * 1323.5

KE(rot) = 330.88 J

6 0
3 years ago
How long does it take for a Ford Econoline van moving at 39.5 m/s to travel 600 m?
Andrej [43]

Answer: B. 15.2s

Explanation: 600/39.5 = about 15.2

6 0
3 years ago
Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
scoundrel [369]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

   Pressure at sea level = 1 atm = 101300 Pa

   we know

   P = ρ g h

   h = \dfrac{P}{\rho\ g}

   h = \dfrac{101300}{1.3\times 9.8}

          h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

  at x = 0  air density = 0

  at x= h   ρ_l = ρ_sl

 assuming density is zero at x - distance

 \rho_x = \dfrac{\rho_{sl}}{h}\times x

now, Pressure at depth x

dP = \rho_x g dx

dP = \dfrac{\rho_{sl}}{h}\times x g dx

integrating both side

P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx

P =\dfrac{\rho_{sl}\times g h}{2}

 now,

h=\dfrac{2P}{\rho_{sl}\times g}

h=\dfrac{2\times 101300}{1.3\times 9.8}

  h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

6 0
3 years ago
A 46.0-kg box is being pushed a distance of 8.80 m across the floor by a force P whose magnitude is 171 N. The force P is parall
dolphi86 [110]

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

W_P=Px

P: applied force = 171N

x: distance in which the for P is applied = 8.80m

you replace the values of P and x and obtain:

W_P=(171N)(8.80m)=1504.8J

(b) The work don by the friction force is:

W_f=F_fx=\mu N x=\mu Mg x

μ = coefficient of kinetic friction = 0.250

M: mass of the box = 46.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J

(c) The Normal force is

N=Mg=(46.0kg)(9,8m/s^2)=450.8N

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

W_N=0J

(d) the same as before:

W_g=0J

8 0
2 years ago
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