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3 years ago

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

1 answer:

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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If the starting population of 5 rabbits grows at 200% each year, how many will there be 50 years?

just olya [345]

Answer:

Alot of god dam rabbits

Explanation:

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3 years ago

Read 2 more answers

In part B of the lab, when the current flows through the orange part of the wire from right to left, the wire deflects (or moves

s2008m [1.1K]

Answer:

This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.

We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:

F = L*(IxB)

if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:

I = i*(1, 0, 0)

And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.

To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.

Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.

For example, if the magnetic field is in the positive z-axis, we will point upwards.

Now the palm of our hand tells us in which direction the force is applied.

This is the right-hand rule.

For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.

7 0

3 years ago

6 An object can never have

MAVERICK [17]

Answer:

Explanation:

Speed is the magnitude of the velocity vector, so it can never be negative.

6 0

3 years ago

A ship is travelling due east at 30 km/hr and a boy runs across the deck

dsp73

Answer:

Vr = 20 [km/h]

Explanation:

In order to solve this problem, we have to add the relative velocities. We must remember that velocity is a vector, therefore it has magnitude and direction. We will take the sea as the reference measurement level.

Let's take the direction of the ship as positive. Therefore the boy moves in the opposite direction (Negative) to the reference level (the sea).

$V_{r}=30-10\\V_{r}=20 [km/h]$

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3 years ago

I was driving along at 20 m/s, trying to change a CD and not watching where I was going. When I looked up, I found myself 45 m f

cricket20 [7]

Answer:

$a=4.44\frac{m}{s^2}$

Explanation:

First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

$t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s$

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.

Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

$d=vt\\d=20\frac{m}{s}(0.5s)=10m$

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:

$d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}$

7 0

4 years ago