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4 years ago

Density is ———- Through the world’s ocean

Physics

1 answer:

Montano1993 [528]4 years ago

7 0

Density is varying through the world’s oceans. The density of ocean water is not uniform.

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A solid conducting sphere of radius 2.00 cm has a charge of 6.77 μC. A conducting spherical shell of inner radius 4.00 cm and ou

madreJ [45]

Answer:

Part a)

E = 0

Part b)

$E = 6.77 \times 10^7 N/C$

Part c)

Electric field inside the conductor is again zero

$E = 0$

Part d)

$E = 8.52 \times 10^6 N/C$

Explanation:

Part a)

conducting sphere is of radius

R = 2 cm

so electric field inside any conductor is always zero

So electric field at r = 1 cm

E = 0

Part b)

Now at r = 3 cm

By Gauss law

$E = \frac{kq}{r^2}$

$E = \frac{(9\times 10^9)(6.77 \muC)}{0.03^2}$

$E = 6.77 \times 10^7 N/C$

Part c)

Again when we use r = 4.50 cm

then we will have

Electric field inside the conductor is again zero

$E = 0$

Part d)

Now at r = 7 cm

again by Gauss law

$E = \frac{kQ}{r^2}$

$E = \frac{(9\times 10^9)(6.77\mu C - 2.13\mu C)}{0.07^2}$

$E = 8.52 \times 10^6 N/C$

5 0

4 years ago

Question 2 (1 point)

mixer [17]

Answer:

The initial vertical velocity is zero, u = 0 m/s

Explanation:

Given;

height of the table, h = 0.55 m

horizontal distance traveled by the tennis, x = 0.12 m

Apply the following kinematic equation;

h = ut + ¹/₂gt²

where;

u is the initial vertical velocity = 0, since the tennis ball rolled off the edge of a table.

h = ¹/₂gt²

The time to fall from the vertical height is given by;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2(0.55)}{9.8} }\\\\t = 0.335 \ s$

The initial horizontal velocity of the tennis is given by;

x = vₓt

vₓ = x / t

vₓ = (0.12) / (0.335)

vₓ = 0.358 m/s

Therefore, the initial vertical velocity is zero, u = 0 m/s and initial horizontal velocity, vₓ is 0.358 m/s

3 0

4 years ago

If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her

Marysya12 [62]

Answer:

μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0

fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0

fr = cotan θ (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

F1 - fr = 0

F1 = fr

the friction force has the expression

fr = μ N

Y Axis

N - W - W_painter = 0

N = W + W_painter

we substitute

fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)

μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values they give

μ = cotan θ (12/2 + 0.3 55) / (12 + 55)

μ = cotan θ (22.5 / 67)

μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

cotan 45 = 1 / tan 45 = 1

the result is

μ = 0.336

5 0

4 years ago

In a closed ring a and in an open ring magnet are falling along the x axis of thrme ring .the current generated in a and b have

zhenek [66]

Answer:

ALL AWNERS HERE

Explanation:

https://quizlet.com/449884025/test-3-physics-2-flash-cards/

8 0

3 years ago

Which quantity can be calculated using Newton’s second law of motion?

DiKsa [7]

Newton's 2nd law of motion:

Net Force = (mass) x (acceleration) .

The law shows the relationship among an object's mass
and acceleration, and the net force acting on it.

If you know any two of the quantities in the formula,
the law can be used to calculate the third one.

3 0

3 years ago