All categories

3 years ago

Help!!! Pleaaaaaseee SOS!!

Physics

1 answer:

OLga [1]3 years ago

7 0

Answer:

D. When its weight is equal to the buoyant force exerted by the water.

Explanation:

A fluid exerts an upward force on a body submerged in it. This force that tends to lift the body is called floating force.

The flotation force that acts on a body submerged in a fluid is equal to the weight of the fluid displaced by the body and acts upward passing through the centroid of the displacemented volume.

You might be interested in

Which is the most accurate name for the covalent compound P2O3?

dexar [7]

P2o3 is <span>diphosphorus trioxide.
Add water and you get phosphoric acid</span>

6 0

3 years ago

Read 2 more answers

Research on the Big Five has indicated that approximately _____ percent of traits appear to be inherited.

Evgen [1.6K]

D 50
Hopefully this helps!

4 0

3 years ago

Read 2 more answers

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

8 0

3 years ago

How should the distance, D, between two point charges, q1 and q2, be changed to double their electric potential energy

Licemer1 [7]

The electrostatic potential energy, U, of one point charge q at position d in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force to bring it from the reference position d to that position

$u \: = \: \frac{kq1q2}{d}$

Thus, to double the electric potential energy U we need to reduce the distance of separation by half (1/2) because they are inversely proportion

6 0

3 years ago

Please help I literally don’t understand

Veseljchak [2.6K]

Answer:

A= 2

B=3

C=4

D=5

E=7

F=8

H=12

7 0

3 years ago