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julia-pushkina [17]

3 years ago

13

A steel beam is rotated in a horizontal plane to provide the centripetal acceleration for

1 answer:

oksian1 [2.3K]3 years ago

7 0

Answer:

12 m/s

Explanation:

a = v² / r

78 m/s² = v² / 2.0 m

v = 12.49 m/s

Rounding to two significant figures, the speed is 12 m/s.

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What type of energy is involved when a river moves sediment and erodes its banks?

ziro4ka [17]

<span>First question: The type of energy involved when a river moves sediment and erodes its banks is: option d. Kinetic energy. Kinetic energy is the energy associated with motion. A body (in this case the water) that moves has an energy associated with its motion that is proportional to the speed (exactly to the square of the speed). When the water collides with the banks it is the kinetic energy of the river that erodes it Second question: the answer is the option d. As gravity pulls water down a slope potential energy changes to knietic energy. This is the, water loses altitude and gains velocity. The potential energy. which is proportional to the height, decreases and the kinetic energy, which is proportional to the square of the speed, increases.</span>

8 0

3 years ago

Applied force is the force of support exerted by an object that holds up another object.

kenny6666 [7]

False, applied force is when a person or an object pushes on another object

6 0

3 years ago

Read 2 more answers

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

7 0

3 years ago

If 1.00 mol of argon is placed in a 0.500-L container at 28.0 ∘C , what is the difference between the ideal pressure (as predict

Rudik [331]

Answer:

1.98 atm

Explanation:

Given that:

Temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28 + 273.15) K = 301.15 K

n = 1

V = 0.500 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

P × 0.500 L = 1 ×0.0821 L atm/ K mol × 301.15 K

⇒P (ideal) = 49.45 atm

Using Van der Waal's equation

$\left(P+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$

R = 0.0821 L atm/ K mol

Where, a and b are constants.

For Ar, given that:

So, a = 1.345 atm L² / mol²

b = 0.03219 L / mol

So,

$\left(P+\frac{1.345\times \:1^2}{0.500^2}\right)\left(0.500-1\times 0.03219\right)=1\times 0.0821\times 301.15$

$P+\frac{1.345}{0.25}=\frac{24.724415}{0.46781}$

$P=\frac{24.724415}{0.46781}-\frac{1.345}{0.25}$

⇒P (real) = 47.47 atm

Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm

4 0

2 years ago

What is the difference between collinear and coplanar

DIA [1.3K]

Answer:

Collinear is when points are in the same straight line but coplanar is when points are all on the same plane.

4 0

2 years ago