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3 years ago

What is buoyancy? How is it related to the force of gravity?

Physics

1 answer:

boyakko [2]3 years ago

5 0

Answer:

buoyancy is the ability to float on water or other liquids. this relates to the force of gravity because the stronger that force, the more likely it is to sink.

Explanation:

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Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30

beks73 [17]

Answer:

T=575.16K

Explanation:

To solve the problem we proceed to use the 1 law of diffusion of flow,

Here,

$J=-D\frac{\Delta C}{\Delta x}$

$\Delta C$ is the rate in concentration

$\Delta x$ is the rate in thickness

D is the diffusion coefficient, where,

$D= D_0 exp(\frac{Q_d}{RT})$

Replacing D in the first law,

$J=-(D_0 exp(\frac{-Q_D}{RT}))\frac{\Delta }{\Delta x}$

clearing T,

$T=\frac{Q_d}{R*ln(\frac{J*\Delta x}{D_0*\Delta C})}$

Replacing our values

$T=-\frac{80000}{8.31*ln(\frac{(6.2*10^{-7})(-15*10^{-3})}{(1.43*10^{-9})(0.65-0.30)})}$

$T=-\frac{80000}{-138.09}$

$T=575.16K$

4 0

3 years ago

Can someone give me tips for science camp!! Please

Kamila [148]

Try to have a calm morning before camp. A good night’s sleep and a good breakfast. Make sure to be cautious, follow all the rules for certain areas ( some maybe restricted ). Take lots of photos doing wacky stuff! Learn but have fun learning

5 0

2 years ago

Read 2 more answers

Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up

Mademuasel [1]

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

$v = \sqrt{2as}$

$v = \sqrt{2*18.9*391}$

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

$V= \sqrt{v^2 + 2aS'}$

$V = \sqrt{121.57^2+ 2 (-9.92)(332)}$

V = 90.51 m/s

6 0

3 years ago

Where in the motion is the magnitude of the force from the spring on the object zero? Where in the motion is the magnitude of th

kkurt [141]

Answer:

1. The magnitude of the force from the spring on the object is zero on Equilibrium.

2. The magnitude of the force from the spring on the object is a maximum on The top and bottom.

3. The magnitude of the net force on the object is zero on The Bottom.

4. The magnitude of the force on the object is a maximum on the Top.

Explanation:

1. Because the change in position delta X is zero.

2. Because of delta X.

3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.

4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.

8 0

3 years ago

What is the approximate terminal velocity of a sky diver before the parachute opens

anzhelika [568]

Answer:

The approximate terminal velocity of a sky diver before the parachute opens is 320 km/h.

Explanation:

The terminal velocity is the maximum magnitude of velocity that is attained by the diver when he or she falls in the air.
The terminal velocity of the person diving in air before opening parachute is 320 km/h that means the velocity when the person is experiencing free fall is 320 km/h.
During terminal velocity, we can represent mathematical equation as;

Buoyancy force + drag force = Gravity

6 0

3 years ago