All categories

2 years ago

According to Archimedes' Principle, what condition has to be met for an object to float?you will get branliest

Physics

1 answer:

MissTica2 years ago

3 0

Explanation:

The buoyant force must be greater to float, otherwise it would sink, its like a barrel in water, the more water weight in it the more it sinks, the more air weight the more it rises.

You might be interested in

Evaporating water produces gaseous water. What has occurred during this process?

Katyanochek1 [597]

Answer:

Heat has accelerated water atoms enough to break the surface tension which leads the liquid to turn into a gas

Explanation:

The state of a substance depends on the distribution of its atoms, therefore any atmosphere change (in this case heat) enough to change the atoms Distribution results in a change of state.

brainliest please ;)

3 0

3 years ago

A cheetah can accelerate from rest at the rate of 4m /s

lesya [120]

Acceleration of cheetah (a) = 4m/s²
time = 10s
initial velocity(u) = 0
final velocity = v
distance travelled = s

v = u +at = 0 + 10×4 = 40m/s
s = (v²-u²)/2a = 40²/(2×4) = 1600/8 = 200m

6 0

3 years ago

Is it possible that a speed of 254 and a speed of 100 could be the same speed?

EastWind [94]

Not if both speeds are in the same units.

However, if the 254 is 'centimeters per time' and the 100 is 'inches per time',
then the speeds are equal.

4 0

3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

3 years ago

a rocket, initially at rest, steadily gains speed at a rate of 13.0m/s^2 for 6.40 during take-off. How far did the rocket travel

Aliun [14]

Explanation:

sbsbbsbshdi she sgebsbwhwjge hsshhdhsshshsuus she's used ydhdyuyddhdhdy hehehe due susshsuhdhdhd ffujfufudhdndndjd d

3 0

3 years ago