In each cylinder, draw the piston and the gas particles based on the temperature shown.

To measure specific heat, the student flows air with a velocity of 20 m/s and a temperature of 25C perpendicular to the length

san4es73 [151]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

3 0

3 years ago

The hubbles telescopes orbit is 5.6 x10 ^5 meters above earths suface. the telescope has a mass os 1.1 x10^4 kilograms. earth ex

Andru [333]

(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg

Fam

7 0

3 years ago

How much energy will a stock tank heater rated at 1790 Watts use in a 24 hour period? 1. 1790 × 24 × 3600 Joules 2. 1790 × 24 ×

Rashid [163]

<h2>Option 1 is the correct answer.</h2>

Explanation:

Power of heater, P = 1790 W

Time used, t = 24 hours = 24 x 60 x 60 = 24 x 3600 s

We have the equation

$\texttt{Power}=\frac{\texttt{Energy}}{\texttt{Time}}$

We need to find energy,

Substituting

$\texttt{Power}=\frac{\texttt{Energy}}{\texttt{Time}}\\\\1790=\frac{\texttt{Energy}}{24\times 3600}$

Energy = 1790 x 24 x 3600 J

Option 1 is the correct answer.

3 0

3 years ago

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

GuDViN [60]

Answer:

E/4

Explanation:

The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

Where;

E is the electric field

σ is the surface charge density

ε₀ is the electric constant.

Formula to calculate σ is;

σ = Q/A

Where;

Q is the total charge of the sheet

A is the sheet's area.

We are told the elastic sheet is a square with a side length as d, thus ;

A = d²

So;

σ = Q/d²

Putting Q/d² for σ in the electric field equation to obtain;

E = Q/(2ε₀d²)

Now, we can see that E is inversely proportional to the square of d i.e.

E ∝ 1/d²

The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.

From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;

E_new = E/4

3 0

3 years ago

A girl drops a stone from the top a tower 45m tall. At the same time, a boy standing at the base of the tower, projects another

Advocard [28]

Answer:

(i) The stones meet at 1.8 second

(ii) The point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.

Explanation:

(i)

First we consider the stone dropped by the girl. We have data:

Vi = Initial Velocity of Stone = 0 m/s (Since the stone was initially at rest)

t = Time Period

g = 10 m/s²

s₁ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₁ = Vi t + (0.5)gt²

s₁ = (0)(t) + (0.5)(10)t²

s₁ = 5t² ----- equation (1)

Now, we consider the stone throne vertically upward by the boy. We have data:

Vi = Initial Velocity of Stone = 25 m/s

t = Time Period

g = - 10 m/s² (negative sign due to upward motion)

s₂ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₂ = Vi t + (0.5)gt²

s₂ = (25)(t) + (0.5)(-10)t²

s₂ = 25t - 5t² ----- equation (2)

At, the point where both the stones meet, the sum of distances covered by both stones must be equal to the height of building (i.e 45 m).

s₁ + s₂ = 45

using values from equation (1) and equation (2)

5t² + 25t - 5t² = 45

25t = 45

t = 45/25

(ii)

using this value of of t in equation (2)

s₂ = (25)(1.8) - (5)(1.8)²

using this value of of t in equation (1)

s₁ = (5)(1.8)²

<u>Hence, the point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.</u>

6 0

3 years ago