I Need help please with this question..

In which point the electric intensity of a sphere is maximum

g100num [7]

Here, as the charge is uniformly distributed in the sphere, we will consider s as an area of the sphere which is, s=4πr2 and r is radius of the gaussian surface shown in the figure above. From this, it can be seen that

7 0

2 years ago

In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 1850 J of work is done on the gas

hammer [34]

Answer:

The value of change in internal energy of the gas = + 1850 J

Explanation:

Work done on the gas (W) = - 1850 J

Negative sign is due to work done on the system.

From the first law we know that Q = Δ U + W ------------- (1)

Where Q = Heat transfer to the gas

Δ U = Change in internal energy of the gas

W = work done on the gas

Since it is adiabatic compression of the gas so heat transfer to the gas is zero.

⇒ Q = 0

So from equation (1)

⇒ Δ U = - W ----------------- (2)

⇒ W = - 1850 J (Given)

⇒ Δ U = - (- 1850)

⇒ Δ U = + 1850 J

This is the value of change in internal energy of the gas.

7 0

3 years ago

A cube made Of an unknown material has a height of 9 symeters the mass of this cube is 3645 g. calculate the density of this cub

kherson [118]

Answer: $5 \frac{g}{cm^{3}}$

Explanation:

The density $\rho$ of a material is given by:

$\rho=\frac{m}{V}$ (1)

Where:

$m=3645 g$ is the mass of the cube

$V$ is the volume of the cube

Now, the volume of a cube is equal to the length $L$ of its edge to the power of 3:

$V=L^{3}$ (2)

If we know $L=9 cm$ , the volume of this cube is:

$V=(9 cm)^{3}=729 cm^{3}$ (3)

Substituting (3) in (1):

$\rho=\frac{3645 g}{729 cm^{3}}$ (4)

$\rho=5 \frac{g}{cm^{3}}$ This is the density of the cube

8 0

3 years ago

Assume the space shuttle's main engines produce 764,576 newtons of thrust, and the shuttle has a mass of 78,018 kg. Why does the

Nady [450]

Weight of anything = (mass) x (gravity in the place where the thing is)

Weight of anything on Earth = (mass) x (9.81 m/s²)

Weight of the shuttle = (78,018 kg) x (9.81 m/s²)

Weight of the shuttle, on Earth = 765,357 Newtons

Thrust of main engines = 764,576 Newtons

Are you starting to see the problem yet ?

The weight of the whole thing standing on the launch pad is 751 Newtons more than the maximum thrust of the main engines, and the engines can't lift it ! Even with all throttles wide open, the main engines alone would need about 175 more pounds of thrust to budge that load off the ground. Even with the pedal to the metal, with flame and smoke belching out and covering the whole launch complex, the shuttle would just sit there and never leave the pad.

Well, no. That's not exactly what would happen. As the fuel in the main monster fuel tank is burned, the weight decreases. So it would actually happen like this: After the man announced "Zero ! We have ignition ! All engine running !", the ship would just sit there on the pad ... at first. It would go nowhere and not even wiggle, UNTIL the first 175 pounds of fuel got burned without accomplishing anything. The ship would then be 175 pounds lighter. At that point, the weight would be exactly equal to the thrust of the main engines, and the vertical forces on the ship would be balanced. Then, as MORE fuel continued to be wasted and the weight continued to decrease, the main engines could just begin to lift the ship off the pad.

So the correct answer is choice-D . It tells the whole story, quicker than I can tell it.

4 0

3 years ago

Charge is flowing through a conductor at the rate of 420 C/min. If 742 J. of electrical energy are converted to heat in 30 s., w

Y_Kistochka [10]

Answer:

3.53 V

Explanation:

Electric charge: The is the rate of flow of electric charge along a conductor.

The S.I unit of electric charge is C.

Mathematically it is expressed as,

Q = It ............................ Equation 1

Where Q = electric charge, I = current, t = time.

I = Q/t.......................... Equation 2

From the question, charge flows through the conductor at the rate of 420 C/mim

Which means in 1 min, 420 C of charge flows through the conductor.

Hence,

Q = 420 C, t = 1 min = 60 seconds

Substitute into equation 2

I = 420/60

I =7 A

Also

P = VI......................... Equation 3

Where P = power, V = potential drop, I = current.

V = P/I................... Equation 4

Note: Power = Energy/time

From the question, P = 742/30 = 24.733 W. and I = 7 A.

Substitute these values into equation 4

V = 24.733/7

V = 3.53 V

Hence the potential drop across the conductor = 3.53 V

4 0

3 years ago