Planet-solar system-galaxy-local group. what comes next

Consider a cube with one corner at the origin and with sides of length 50 cm positioned along the x y z axes. There is an electr

lara [203]

Answer:

Charge in the cube = Q = (2.323 × 10⁻¹⁰) C

Explanation:

E → = < 45, 210 y , 0 > N/C

The length of the cube = 50 cm = 0.8 m

To find the charge in the cube, we need the net flux in the cube

In the z-direction, the electric field is 0, no, net flux in that direction,

In the x-direction, the electric field is constant, the same flux thay enters the face at x = 0 is the same that leaves at x= 0

50 cm, hence, no net flux in that direction too.

But the y direction, the electric field changes according to 210y from y = 0 to y = 0.50 m.

At y = 0, electric field = 210(0) = 0 N/C

At y = 0.50, electric field = 210(0.5) = 105 N/C.

Electric flux = Φ = E A = 105 (0.5 × 0.5) = 26.25 Vm or N/m²C

(the area of the xz plane that the field in this direction passes through is indeed 0.5²)

But according to Gauss' law, the net flux in a box is given by

Φ = Q/ε₀

where Q = charge in the box = ?

ε₀ = (8.85 × 10⁻¹²) C²N·m².

Q = Φε₀ = 26.25 × 8.85 × 10⁻¹²

Q = (2.323 × 10⁻¹⁰) C

Hope this Helps!!!

4 0

4 years ago

yKpoI14uk [10]

D if the blue car started higher it would have more energy but since the red car is lower it is going faster because it’s going down a hill

8 0

3 years ago

A heavy solid disk rotating freely and slowed only by friction applied at its outer edge takes 120 seconds to come to a stop.

alisha [4.7K]

Answer:

The time is 16 min.

Explanation:

Given that,

Time = 120 sec

We need to calculate the moment of inertia

Using formula of moment of inertia

$I=\dfrac{1}{2}MR^2$

If the disk had twice the radius and twice the mass

The new moment of inertia

$I'=\dfrac{1}{2}\times2M\times(2R)^2$

$I'=8I$

We know,

The torque is

$\tau=F\times R$

We need to calculate the initial rotation acceleration

Using formula of acceleration

$\alpha=\dfrac{\tau}{I}$

Put the value in to the formula

$\alpha=\dfrac{F\times R}{\dfrac{1}{2}MR^2}$

$\alpha=\dfrac{2F}{MR}$

We need to calculate the new rotation acceleration

Using formula of acceleration

$\alpha'=\dfrac{\tau}{I'}$

Put the value in to the formula

$\alpha=\dfrac{F\times R}{8\times\dfrac{1}{2}MR^2}$

$\alpha=\dfrac{2F}{8MR}$

$\alpha=\dfrac{\alpha}{8}$

Rotation speed is same.

We need to calculate the time

Using formula angular velocity

$\Omega=\omega'$

$\alpha\time t=\alpha'\times t'$

Put the value into the formula

$\alpha\times120=\dfrac{\alpha}{8}\times t'$

$t'=960\ sec$

$t'=16\ min$

Hence, The time is 16 min.

5 0

3 years ago

A person walks 45 m East and then walks 39 m at an angle 40◦ North of East. What is the magnitude of the total displacement? Ans

vredina [299]

Answer:

r = 78.95 m

Explanation:

given,

Person Walk in east, r₁ = 45 m

Then he walk, r₂= 39 m in the direction of 40◦ North of East.

writing the total displacement in x- direction

r₁ₓ = 45 m

r₂ₓ = 39 cos 40°

r₂ₓ = 29.87 m

total displacement in x-direction

rₓ = r₁ₓ + r₂ₓ

rₓ = 45 + 29.87

rₓ = 74.87 m

displacement in y-direction

r_{1y} = 0 m

r_{2y} = 39 sin 40°

r_{2y} = 25.07 m

total displacement in y- direction

r_y = 25.07 m

using Pythagoras theorem for the magnitude calculation

$r = \sqrt{r_x^2 + r_y^2}$

$r = \sqrt{74.87^2 +25.07^2}$

r = 78.95 m

The magnitude of total displacement is equal to 78.95 m.

8 0

3 years ago

7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr

Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

$d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m$

$d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m$

From the continuity equations in pipes we have to

$A_1V_1 = A_2 V_2$

Where,

$A_{1,2}$ = Cross sectional Area at each section

$V_{1,2}$ = Flow Velocity at each section

Then replacing we have,

$(\pi r_1^2) v_1 = (\pi r_2^2) v_2$

$(1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2$

$v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1$

From Bernoulli equation we have that the change in the pressure is

$\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)$

$7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)$

$7300= 8919.01 v_1^2$

$v_1 = 0.9m/s$

Therefore the speed of flow in the first tube is 0.9m/s

6 0

4 years ago