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hichkok12 [17]
2 years ago
15

Three balls are in water. Ball 1 floats, with half of it exposed above the water level. Ball 2, with a density less than the den

sity of water is held below the surface by a cord anchored to the bottom of the container, so that it is fully submerged. Ball 3, of same radius as ball 2, but of greater mass, is suspended from a rope so that it is fully submerged.
Which is true for Ball 1? Choose the best answer.

1) The magnitude of the buoyancy force is exactly one half of that of the ball's weight.
2) The magnitude of the buoyancy force is more than that of the ball's weight.
3) The magnitude of the buoyancy force is less than that of the ball's weight.
4) The magnitude of the buoyancy force is equal to that of the ball's weight.
Physics
2 answers:
goblinko [34]2 years ago
7 0

Answer:

4) The magnitude of the buoyancy force is equal to that of the ball's weight

Explanation:

Since ball 1 floats without sinking then it is in equilibrium. This means that the net downward force (weight) is balanced by the net upward force (buoyancy forces).

Ymorist [56]2 years ago
4 0

Answer:

The magnitude of buoyancy force is equal to that of ball's weight.

Explanation:

Ball 1 is floating on water. Weight of ball 1 is Fg=m1g  is acting vertically downward

Force of buoyancy FB = ρVdisg is acting vertically upward.

Net force acting on the ball is zero, FB=Fg

Answer

The magnitude of buoyancy force is equal to that of ball's weight.

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Planet A has mass 3M and radius R, while Planet B has mass 4M and radius 2R. They are separated by center-to-center distance 8R.
Aleksandr-060686 [28]

Answer:

Explanation:



In Newton's law of universal gravitation

F = Gm₁m₂/r²

Where G is a gravitational constant = 6.674e-11m³/kgs²

m₁ and m₂ are the masses of the two bodies or objects in question, in kilogram (kg)

r is the distance in meters between them

From the question, the rock is placed halfway between the planets

So, it's distance from planet A is 8R/2 = 4R

And it's distance from planet B is also 8R/2 = 4R

Using F = Gm₁m₂/r²

To Planet A

r = 4R,

m₁ = mass of Rock = m

m₂ = mass of planet A = 3M

So Fa = G mm₂/r² = Gm(3M) / (4R)²

To Planet B,

r = 4R,

m₁ = mass of Rock = m

m₂ = mass of planet B = 4M

Fb = G mm₂/r² = Gm(4M) / (4R)²

Comparing both forces together, we realise that Planet B has the largest force,

so take we F = Fb – Fa

F = Fb – Fa = Gm(4M) / (4R)² – Gm(3M) / (4R)²

F = GmM/16R²)(4–3)

F = GmM/16R²

Note that Force = Mass * Acceleration

So, F = ma

So, ma = GmM/16R² ------- Divide through by m

a = GM/16R²

From the question

M = 7.3×10^23kg

R = 5.8×10^6 m

So, a = (6.674 * 10^-11 * 7.3×10^23)/16(5.8×10^6)²

a = (48.7202 * 10^12)/16(33.64 * 10^12)

a = (48.7202 * 10^12)/(538.4 * 10^12)

a = 48.7202/538.4

a = 0.090517612960760

a = 0.091m/s² ----------Approximated

5 0
2 years ago
A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to
Ne4ueva [31]

Answer:

a) r=4.24cm d=1 cm

b) Q=5x10^{-10} C

Explanation:

The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field

V=1.00x10^{2}v\\E=1.00x10^{4} \frac{N}{C}

V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m

The distance must be the separation the r distance can be find also using

C=\frac{Q}{V_{ab}}

But now don't know the charge these plates can hold yet so

a).

d=0.01m

C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}

A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}

A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } }  \\r=42.55x^{-3}m

b).

C=\frac{Q}{V_{ab}}

Q=C*V\\Q=5x10^{-12} F*1x10^{2}\\Q=5x10^{-10}C

8 0
2 years ago
A 1-kg mass is dropped from a third floor window. The acceleration of the mass is found to be 8 m/s2. What is the average force
Paha777 [63]
Summary:
m=1kg
a=8 m/s^2
g= 9,8 m/s^2
F(ar)=?


I hope to help you

7 0
2 years ago
The position of a 50 g oscillating mass is given by x(t)=(2.0cm)cos(10t−π/4), where t is in s. If necessary, round your answers
Leona [35]
For the answer to the question above, this is the maximum displacement, the spring has only elastic potential energy.
spring is constant @ 5 N/m
maximum displacement = 2 cm = 0.02 m
elastic potential energy = 1/2 kx²
= 0.5 x 5 x 0.02²
So the answer would be
= 0.001 Joules
5 0
2 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m
r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}

here we have

q_1 = 1.6 \times 10^{-19} C

q_2 = -1.6 \times 10^{-19} C

r_1 = r_2 = 1 m

now we have

V = \frac{Ke}{r} - \frac{Ke}{r}

V = 0

Now work done to move another charge from infinite to origin is given by

W = q(V_f - V_i)

here we will have

W = e(0 - 0) = 0

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}

U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})

U = 1.15\times 10^{-30}

Now we know

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2

KE = 4.55 \times 10^{-27} kg

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0
2 years ago
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