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3 years ago

This diagram shows a process the power stars. This process is called

Physics

2 answers:

goldenfox [79]3 years ago

8 0

its c........................................................................

Damm [24]3 years ago

6 0

I think its oxidation.

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A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now

Vesnalui [34]

Answer:

a. 8p

Explanation:

We are given that

Radius of hollow sphere , R1=R

Density of hollow sphere= $\rho$

After compress

Radius of hollow sphere, R2=R/2

We have to find density of the compressed sphere.

We know that

$Density=\frac{mass}{volume}$

$Mass=Density\times volume=Constant$

Therefore, $\rho_1 V_1=\rho_2V_2$

Volume of sphere= $\frac{4}{3}\pi r^3$

Using the formula

$\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3$

$\rho R^3=\rho_2\times \frac{R^3}{8}$

$\rho_2=8\rho$

Hence, the density of the compressed sphere= $8\rho$

Option a is correct.

7 0

2 years ago

The law of reflection establishes a definite relationship between the angle of incidence of a light ray striking the boundary be

sladkih [1.3K]

Answer:

The law of reflection states that the angle of incidence = the angle of reflection.

Explanation:

Reflection is the phenomenon that occurs when a ray of light hits the boundary between two media and it is reflected back into the first medium.

In such a situation, we call:

- angle of incidence: it is the angle between the direction of the incident ray and the normal to the surface

- angle of reflection: it is the angle between the direction of the reflected ray and the normal to the surface

There is a precise relationship between the angle of incidence and the angle of reflection. In fact, the Law of Reflection states that:

- The incident ray, the reflected ray and the normal to the surface all lie within the same plane

- The angle of reflection is equal to the angle of incidence

4 0

3 years ago

a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi

zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

v = u + at

a = (v - u) / t

Substituting the given values in the above equation

a = (10 - 0) / 3

= 3.33 m/s²

Using the second equations of motion

s = ut + 1/2 at²

= 0 + 0.5 x 3.33 x 3²

= 15 m

The force exerted by the car

F = m x a

= 1500 Kg x 3.33 m/s²

= 5000 N

The work done by the car,

W = F x S

= 5000 N x 15 m

= 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

P = W / t

= 75,000 J / 3 s

= 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0

3 years ago

Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charg

nata0808 [166]

Answer:

a) $U_{e} = 3 \times 10^{10}\,J$ , b) $v \approx 7745.967\,\frac{m}{s}$

Explanation:

a) The potential energy is:

$U_{e} = Q \cdot \Delta V$

$U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)$

$U_{e} = 3 \times 10^{10}\,J$

b) Maximum final speed:

$U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }$

The final speed is:

$v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }$

$v \approx 7745.967\,\frac{m}{s}$

3 0

3 years ago

How long does it take a 22 kW steam engine to do 5.6 × 107 J of work? Answer in units of s.

Lena [83]

<h2>Time needed is 2545.45 seconds.</h2>

Explanation:

We know equation for power

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

Here we need to find time when for a 22 kW steam engine to do 5.6 × 10⁷ J of work.

Work = 5.6 × 10⁷ J

Power = 22 kW = 22 x 10³ W

Substituting

$\texttt{Power = }\frac{\texttt{Work}}{\texttt{Time}}\\\\22\times 10^3=\frac{5.6\times 10^7}{\texttt{Time}}\\\\\texttt{Time}=2545.45s$

Time needed is 2545.45 seconds.

5 0

3 years ago