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2 years ago

What is the approximate mass of air in a living room 4.5m×3.4m×2.9m? the density of air is 1.29 kg/m3?

1 answer:

6 0

First we have to calculate the volume of the living room:
V = L x W x H = 4.5 m * 3.4 m * 2.9 m
V = 44.37 m³
We know that Density = 1.29 kg/m²
D = m / V
m = D · V
m = 1.29 kg/m³ · 44.37 m³
m = 57.2373 kg ≈ 57.2 kg
Answer: The approximate mass of air in living room is 57.2 kg.

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Which material would be best for thawing frozen food quickly? O A. A plastic plate O B. A steel plate O C. A glass plate O D. A

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Answer: a steel. Plate

Explanation: just did it on apex

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3 years ago

If the loop is removed from the field region in a time interval of 2.8 ms , find the average emf that will be induced in the wir

nekit [7.7K]

The question is incomplete. The complete question is :

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.2 T is directed along the positive z-direction, which is upward. (a)If the loop is removed from the field region in a time interval of 2.8 ms, find the average emf that will be induced in the wire loop during the extraction process.

Solution :

Let us consider a $\text{circular loo}p \text{ of wire}$ which has a $\text{radius}$ of r = $15$ cm.

It is oriented horizontally along the xy-plane and is located in the region of an $\text{uniform magnetic field}$ , such that it points in the positive z direction and having a magnitude of B = 1.2 T.

Now if the loop $\text{is removed from the field region}$ in a time interval of Δt = 2.8 ms. Initially the magnetic field and the area points is in the same direction, so that the angle between them is Ф = 0°, thus the initial and the final fluxes are :

$\phi_{B,i}=BA \cos (\phi) = BA$ and $\phi_{B,f} = 0$

Area A = $\pi r^2.$ The induced emf equals to the change in the flux, and is divided by the time that it takes to go from the initial flux, Δt and multiplied by the number of turns N = 1, i.e. ,

$\epsilon = -\frac{\Delta \phi_{B}}{\Delta t}$

$=-\frac{0-(1.2 T)\pi(0.15^2)}{2.8 \times 10^{-3}}$

= 30.27 V

Therefore, the emf generated is 30.27 V.

7 0

2 years ago

A machine part rotates at an angular speed of 0.76 rad/s; its speed is then increased to 2.96 rad/s using an angular acceleratio

son4ous [18]

Answer:

θ = 4.092 °

Explanation:

Given that

Initial Angular speed ,ωi = 0.76 rad/s

Final angular speed ,ωf= 2.96 rad/s

The angular acceleration ,α = 1 rad/s²

We know that

ωf² = ωi ²+ 2 α θ

θ=Angle rotates

2.96² = 0.76²+ 2 x 1 x θ

$\theta=\dfrac{2.96^2-0.76^2}{2}\ degrees\\\theta=4.092\ degrees\\$

Therefore the angle rotates by machine part will be 4.092 degrees.

θ = 4.092 °

5 0

3 years ago

calculate the density of a neutron star with a radius 1.05 x10^4 m, assuming the mass is distributed uniformly. Treat the neutro

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To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.

Mathematically it can be expressed as

$\rho = \frac{m}{V}$