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Sindrei [870]
2 years ago
13

Imagine you are in a small boat on a small pond that has no inflow or outflow. If you take an anchor that was sitting on the flo

or of the boat and lower it over the side until it sits on the ground at the bottom of the pond, will the water level rise slightly, stay the same, or lower slightly?Two students, Ian and Owen, are discussing this. Ian says that the anchor will still displace just as much water when it is sitting on the bottom of the pond as it does when it is in the boat. After all, adding the anchor to the boat causes the water level in the lake to rise, and so would immersing the anchor in the pond. So Ian reasons that both displacements would be equal, and the lake level remains unchanged.
Physics
1 answer:
inn [45]2 years ago
3 0

Answer;

The pond's water level will fall.

Explanation;

Archimedes principle explains that a floating body will displace the amount of water that weighs the same as it, whereas a body resting on the bottom of the water displaces the amount of water that is equal to the body's volume.

When the anchor is in the boat it is in the category of floating body and when it is on the bottom of the pond it is in the second category.

Since anchors are naturally heavy and denser than water, the amount of water displaced when the anchor is in the boat is greater than the amount of water displaced when the anchor is on the bottom of the pond since the way anchors are doesn't make for them to have considerable volume.

When the anchor is dropped to the bottom of the pond, the water level will therefore fall. If the anchor doesn't reach the bottom it is still in the floating object category and there will be no difference to the water level, but once it touches the bottom of the pond, the water level of the pond drops.

Hope this Helps!!!

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Charged beads are placed at the corners of a square in the various configurations shown in
Akimi4 [234]

Answer:

The consecutive charge configuration has a more intense field than alternating

Explanation:

In each corner we place a different account there are only two different settings, see attached.

In the case of alternating charging (+ - + -) see diagram 1, the electric field in the center is canceled in pairs, resulting in a zero field

In the case of consecutive loads (+ + - -) in this case we have a result between the two charges, therefore the total field is

          E = 2 k q / ra2 a cos 45

The consecutive charge configuration has a more intense field than alternating

8 0
2 years ago
A(n) ________ has charge but negligible mass, whereas a(n) ________ has mass but no charge.
vfiekz [6]
Electron;Neutron is the correct answer.
6 0
2 years ago
Read 2 more answers
A flying mosquito hits the windshield of a moving car and gets smashed, but the car is intact. Which of the following statements
zhannawk [14.2K]

Answer:

Action and reaction are the same

Explanation:

Let us carefully analyze the situation in the context of Newton's third law, when the car hits the mosquito exerts a force that acts on the mosquito, ACTION and the mosquito responds to the force that is being applied with a force of equal magnitude and direction or opposite applied to the car (Reaction).

Consequently, the correct answer is: The force of the car on the mosquito is of the same magnitude as that of the mosquito on the car

6 0
2 years ago
I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =
Helen [10]
Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
7 0
2 years ago
When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What
ser-zykov [4K]

Answer:

Electric field, E=4\times 10^5\ N/C

Explanation:

It is given that,

Magnitude of charge, q_o=2\ nC=2\times 10^{-9}\ C

Force experienced, F=8\times 10^{-4}\ N

We need to find the electric field at the origin. It is given by :

F=q_o\times E

E=\dfrac{F}{q_o}

E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}

E=4\times 10^5\ N/C

So, the electric field at the origin is 4\times 10^5\ N/C. Hence, this is the required solution.

3 0
2 years ago
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