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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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The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho=rho0[1+α(T−T0)], where T0
Readme [11.4K]

Answer:

I = ΔVA[1 - α (T₀ - T)]/Lρ₀

Explanation:

We have the following data:

ΔV = Battery Terminal Voltage

I = Current through wire

L = Length of wire

A = Cross-sectional area of wire

T = Temperature of wire, when connected across battery

T₀ = Reference temperature

ρ = Resistivity of wire at temperature T

ρ₀ = Resistivity of wire at reference temperature

α = Temperature Coefficient of Resistance

From OHM'S LAW we know that;

ΔV = IR

I = ΔV/R

but,  R = ρL/A   (For Wire)

Therefore,

I = ΔV/(ρL/A)

I = ΔVA/ρL

but,   ρ = ρ₀[1 + α (T₀ - T)]

Therefore,

I = ΔVA/Lρ₀[1 + α (T₀ - T)]

I = [ΔVA/Lρ₀] [1 + α (T₀ - T)]⁻¹

using Binomial Theorem:

(1 +x)⁻¹ = 1 - x + x² - x³ + ...

In case of [1 + α (T₀ - T)]⁻¹, x = α (T₀ - T).

Since, α generally has very low value. Thus, its higher powers can easily be neglected.

Therefore, using this Binomial Approximation, we can write:

[1 + α (T₀ - T)]⁻¹ = [1 - α (T₀ - T)]

Thus, the equation becomes:

<u>I = ΔVA[1 - α (T₀ - T)]/Lρ₀ </u>

3 0
2 years ago
Lidia plans an experimental investigation to see how the thickness of a lens affects the point where a beam of light is focused.
alexandr1967 [171]

Answer:

The type of light and the material of lenz.

Explanation:

1) As the investigation is based on how the thickness of a lens effect the other variable. Thickness of the lenz is independent variable. So Lidia has to experiment with the different thicknesses in order to find the effect on dependent variable.

2) As the investigation is based to find the point where the beam of light is focused. It's a dependent variable and Lidia has no control over it. So the only thing she can do is to measure and observe how it respond to the changes in independent variable.

3) For conclusion, she has to make sure that the other variables are not effecting the output or results that is the beam point where the light is focused. So she must have to kept constant the type of light and material of lenz otherwise she won't be able to discriminate the effect of thickness of lenz from other causes.

8 0
2 years ago
A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±
vampirchik [111]

Answer: A) 2 B) 4 C) 1

Explanation:

The Electric field from a parallel-plate capacitor  is given by:

A) E=Q/(L^2 * ε0) so if we put a charge double the final electric field is double that the original.

B) from the above expression for the electric field,  If the size of the plate is double, then the E final is four times weaker that the original.

C) If the distante between plates is doubled the final electric field is the same that initial.

3 0
2 years ago
The air in a 6.00 L tank has a pressure of 2.00 atm. What is the final pressure, in atmospheres, when the air is placed in tanks
ser-zykov [4K]

Explanation:

Given that,

Initial volume of tank, V = 6 L

Initial pressure, P = 2 atm

We need to find the final pressure when the air is placed in tanks that have the following volumes if there is no change in temperature and amount of gas:

(a) V' = 1 L

It is a case of Boyle's law. It says that volume is inversely proportional to the pressure at constant temperature. So,

PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{1}\\\\P'=12\ atm

(b) V' = 2500 mL

New pressure becomes :

PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{2500\times 10^{-3}}\\\\P'=4.8\ atm

(c) V' = 750 mL

New pressure becomes :

PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{750\times 10^{-3}}\\\\P'=16\ atm

(d) V' = 8 L

New pressure becomes :

PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{8}\\\\P'=1.5\ atm

Hence, this is the required solution.

3 0
2 years ago
Superman is standing 393 m horizontally away from Lois Lane. A villain drops a rock from 4.00 m directly above Lois. If Superman
Sergio039 [100]

Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 4=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4\times 2}{9.81}}\\\Rightarrow t=0.903\ s

s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2

The acceleration of Superman would be -963.93 m/s² from Lois' perspective

6 0
2 years ago
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