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3 years ago

What is buoyant force?

Physics

1 answer:

Nostrana [21]3 years ago

3 0

This is the upthrust on an object which is placed inside a fluid

This force act upwards and always push upwards

so the correct answer is given as

D. A force within a fluid that pushes upward

this force is always due to pressure difference at two levels of

at lower level since pressure is more that is why the force is upwards and this upthrust is known as Buoyancy

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Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the

BlackZzzverrR [31]

relation between potential difference and electric field is given as

$E . d = \Delta V$

so here we know that

d = 3 cm

$\Delta V = 30 V$

$E \times 0.03 = 30$

$E = 1000 N/C$

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

5 0

3 years ago

Three crates with various contents are pulled by a force F pull = 3615 N across a horizontal, frictionless roller‑conveyor syste

Dmitry_Shevchenko [17]

Answer:

m1=914.9kg

m2=604.9kg

m3=864.75kg

Explanation

I think we are suppose to find the mass of the crate.

The effective force that moves the body in positive x direction is 3615N

ΣFx = Σma

Then Fx=3615N

Then the masses be m1, m2 and m3

Then,

ΣF = Σ(ma)

3615=(m1+m2+m3)a

Given that a=1.516

The masses are

m1+m2+m3=, 2384.56. Equation 1

Between mass 1 and mass 2 is, F12=1387.

The effective force that pull mass 1 is 1387.

F12=m1 ×a

Therefore,

m1=F12/a

m1=1387/1.516

m1=914.9kg.

The effective force that pulls crate 1 and crate 2 is F23

F23=(m1+m2)a

Therefore

2304=(m1+m2)a

Therefore, since a=1.516

m1+m2=2304/1.516

m1+m2=1519.8kg

Since m1=914.9kg

So, m2=1519.8-m1

m2=1519.8-914.9

m2=604.9kg

Also from equation 1

m1+m2+m3=2384.56

Since m1=914.9kg and m2=604.9kg

Then, m3=2384.56-604.9-914.9

m3=864.75kg

3 0

3 years ago

_____ motion occurs when tectonic plates move apart from each other. A. Divergent B. Transform C. Convergent

Likurg_2 [28]

A. Divergent (pretty sure)

4 0

4 years ago

Read 2 more answers

Solve using correct significant figures and indicating maximum absolute uncertainty.

Vera_Pavlovna [14]

We use the criterion of significant figures to find the result with reliable figures

X = 9.2 10-5

now with the propagation of errors we obtain the result with its uncertainty

X ± ΔX = (9.2 ± 0.5) 10⁻⁵

given Parameter

* expression values with their absolute errors

to find

* the result with the correct significant figures

* the absolute error of the expression

Significant figures are defined with the number of decimals that give information, the number of figures in a quantity gives information about the uncertainty of this quantity.

There are two criteria for applying significant figures:

* Add and subtract the result of going with the number of decimal places of the figure that has the least

* Product and division as a result of going with the least number of significant figures than the value that has the least.

Remember that the zero to the left do not form a pair of the significant figures

Let's apply this belief to the case presented, let's write the precaution

$x = \frac{a-b}{c}$

where in this case they are worth

a = 0.0336 ± 0.0002

b = 0.010 ± 0.001

c = 255.4 ± 0.4

We see that the significant figures of each parameterize (a, b, c) and their absolute errors are correct.

Let's apply the criteria to the operation

a-b = 0.0336 - 0.010

a- b = 0.0236

we apply the criterion of significant figures for the subtraction, the result must be with 3 decimal places

a - b = 0.024

let's do the other operation

X = $\frac{a-b}{c}$

X = 0.024 / 255.4

X = 9.24 10⁻⁵

We apply the criterion of significant figures for the division, in this case the result is left with two significant figures

X = 9.2 10⁻⁵

The uncertainty or error of the measurements is of most importance as it determines how many significative figures are reliable at a given magnitude.

If the magnitudes are measured with some type of instrument, the absolute error is given by the appreciation of the instrument, if the magnitude is calculated using some equation, the errors must be propagated using the variations of each parameter in the worst case.

the uncertainty of the calculated quantity (X) is

$\Delta X = | \frac{dX}{da}| \Delta a + | \frac{dX}{db} | \Delta b + | \frac{dX}{dc}| \Delta c$

let's perform the derivatives

$\frac{dX}{da} = \frac{1}{c}$

$\frac{dX}{db} = - \frac{1}{c}$

$\frac{dX}{dc} = - \frac{a-b}{c^2}$

we substitute

remember that the bulk value guarantees that we tune the worst case. So all the mistakes add up

ΔX = $\frac{1}{c}$ Δa + $\frac{1}{c}$ Δb + $\frac{a-b}{c^2}$ Δc

ΔX = $\frac{1}{c}$ (Δa + Δb) + $\frac{a-b}{c^2}$ Δc

we substitute

ΔX = $\frac{1}{255.4}$ (0.0002 + 0.001) + $\frac{0.0336-0.010}{255.4^2}$ 0.4

ΔX = 4.698 10⁻⁶ + 1.45 10⁻⁷

DX = 4.8 10-6

Absolute errors must be given with a single significant figure

ΔX = 5 10⁻⁶

The result of the requested quantity using the criterion of significant figures and propagation of errors is

X ± ΔX = (9.2 ± 0.5) 10⁻⁵

learn more about significative figure here:

brainly.com/question/18955573

8 0

3 years ago

Dan is gliding on his skateboard at 4.00m/s . He suddenly jumps backward off the skateboard, kicking the skateboard forward at 6

frozen [14]

To solve this problem we will apply the concepts related to the conservation of the Momentum. For this purpose we will define the momentum as the product between mass and velocity, and by conservation the initial momentum will be equal to the final momentum. Mathematically this is,

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$