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3 years ago

13

Explain why an object floats on water. Use terms like buoyancy force and gravitational force.

1 answer:

telo118 [61]3 years ago

5 0

Answer:

If an object pushes out an amount of water equal to its own weight, the upward force acting on it will be equal to gravity - and the object will float.

Explanation:

The buoyant force has an impact on the object in the water and equals the weight of the water displaced by the object. Every object placed in water has some buoyancy force that pushes it against the gravitational force, and this means that any object loses weight in the water.

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John wants a new video tape and a new shirt, but he only has $12.00. He buys the video tape and gets $0.52 change. What was his

mestny [16]

Answer:

You have the answer in your comments. I will be copying it so your question doesn't get deleted.

The answers is $0.58

$11.48

the video tape

the new shirt

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3 years ago

17. ____ Objects with more mass have

AnnZ [28]

a. more gravity

This is the answer.

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3 years ago

Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

aev [14]

Answer:

Explanation:

Given

charge of first body $q_1=2.5\ mu C$

charge of second body $q_2=-3.5\ mu C$

Particle 1 is at origin and particle 2 is at $x=0.6\ m$

third Particle which charge +q must be placed left of $2.5\mu C$ because it will repel the q charge while $-3.5\mu C$ will attract it

suppose it is placed at a distance of x m

$F_{1q}=\frac{kq(2.5)}{x^2}$

$F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}$

$F_{1q}+F_{2q}=0$

$\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0$

$\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}$

$\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}$

$0.6+x=1.1832x$

$x=3.27\ m$

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3 years ago

What is the symbol for ammonium

rosijanka [135]

Symbol of ammonium is

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3 years ago

A test car starts from rest on a horizontal circular track of 115-m radius and increases its speed at a uniform rate to reach 90

Wewaii [24]

Answer:

a= 3.49 m/s^2

Explanation:

magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.

we know that tangential acceleration a_t= change in velocity /time taken

now 90 km/h = 25 m/s

a_t = 25/17 = 1.47 m/s^2.

radial acceleration a_r = v^2/r

v= a_t×t = 1.47×13 = 19.11 m/s

a_r = 19.11^2/115= 3.175

now,

$a= \sqrt{a_t^2+a_r^2}$

$a= \sqrt{1.47^2+3.175^2}$

a= 3.49 m/s^2

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3 years ago