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Delvig [45]
3 years ago
7

The buoyant force on an object fully submerged in a liquid depends on (select all that apply)

Physics
1 answer:
Natasha_Volkova [10]3 years ago
8 0
2. The object's volume.
 3. The density of the liquid.  
Remember what the buoyant force is. It's the lifting force caused by the displacement of a fluid. I'm using the word fluid because it can be either a liquid or gas. For instance a helium balloon floats due to the buoyant force exceeding the mass of the balloon. So let's look at the options and see what's correct. 
 1. Object's mass
 * This doesn't affect the buoyant force directly. It can have an effect if the object's mass is lower than the buoyant force being exerted. Think of a boat as an example. The boat is floating on the top of the water. If cargo is loaded into the boat, the boat sinks further into the water until the increased buoyant force matches the increased mass of the boat. But if the density of the object exceeds the density of the fluid, then increasing the mass of the object will not affect the buoyant force. So this is a bad choice. 
 2. The object's volume.
 * Yes, this directly affects the buoyant force. So this is a good choice. 
 3. The density of the liquid.
 * Yes, this directly affects the buoyant force. You can drop a piece of iron into water and it will sink. You could also drop that same piece of iron into mercury and it will float. The reason is that mercury has a much higher density than water. So this is a good choice. 
 4. Mass of the liquid
 * No. Do not mistake mass for density. As a mental exercise, imagine the buoyant force on a small piece of metal dropped into a swimming pool. Now imagine the buoyant force on that same piece of metal dropped into a lake. In both cases, the buoyant force is the same, yet the lake has a far greater mass of water than the swimming pool. So this is a bad choice.
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Current was discovered not invented​
irina1246 [14]

Answer:

that's true current is discoverd originally by benjimin franklin Followed by tesla and Edison.

Explanation:

3 0
3 years ago
A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a
Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C =  50 - x cm

We know that ,electric field is given as

E=K\dfrac{q}{r^2}

K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0
3 years ago
What should #1 be labeled?
aliya0001 [1]
B. North Pole because it’s pointing at the north pole, and not the others
7 0
3 years ago
Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plate
vichka [17]

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F   ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

<u>CB = 4.45 x 10⁻⁹ F = 4.45 nF</u>

5 0
3 years ago
Two motors in a factory are running at slightly different rates. One runs at 825.0 rpm and the other at 786.0 rpm. You hear the
mixas84 [53]

Answer:

T=1.54s

Explanation:

From the question we are told that:

Speed of Motor 1 \omega_1=825rpm=>2 \pi 13.75

Speed of Motor 2 \omega_1=786rpm=>2 \pi 13.1

Therefore

Frequency of Motor 1 f_1=13.75

Frequency of Motor 2  f_2= 13.1

Generally the equation for Time Elapsed is mathematically given by

T=\frac{1}{df}

Where

df=f_1-f_2

df=13.75-13.1

df=0.65Hz

Therefore

T=\frac{1}{65}

T=1.54s

5 0
3 years ago
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