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2 years ago

5

A force of 10.0 newtons acts at an angle 20.0 degrees from vertical. What are the horizontal and vertical components of the forc

e?

1 answer:

erik [133]2 years ago

8 0

Horizontal component = (10N) · sin (20°) = 3.42... N (rounded)

Vertical component = (10N) · cos (20°) = 9.39... N (rounded)

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There have been several proposed atomic models during the last 150 years. Which model best illustrates the Bohr model. This mode

Eva8 [605]

<span>Despite the Quantum Mechanical Model treating the electron mathematically as a wave rather than fixed patterns, the Quantum Mechanical model best illustrates the Bohr model because both models of the atom assign specific energies to an electron.</span>

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2 years ago

Read 2 more answers

Points a and b lie in a region where the y-component of the electric field is Ey=α+β/y2. The constants in this expression have t

Drupady [299]

Answer:

V_{a} - V_{b} = 89.3

Explanation:

The electric potential is defined by

$V_{b} - V_{a}$ = - ∫ E .ds

In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.

V_{b} - V_{a} = - ∫ E ds

We substitute

V_{b} - V_{a} = - ∫ (α + β/ y²) dy

We integrate

V_{b} - V_{a} = - α y + β / y

We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)

V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

V_{b} - V_{a} = - 89.3 V

As they ask us the reverse case

V_{b} - V_{a} = - V_{b} - V_{a}

V_{a} - V_{b} = 89.3

3 0

2 years ago

A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington

Roman55 [17]

Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

The current has velocity vector (relative to the Earth)

$\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath$

The swimmer's resultant velocity (her velocity relative to the Earth) is then

$\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}$

$\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath$

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

$\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ$

which is approximately 41º west of north.

6 0

2 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

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

qaws [65]

Answer:

The distance between both cars is 990 m

Explanation:

The equations for the position and the velocity of an object moving in a straight line are as follows:

x = x0 + v0 * t + 1/2 * a * t²

v = v0 + a * t

where:

x = position of the car at time "t"

x0 = initial position

v0 = initial speed

t = time

a = acceleration

v = velocity

First let´s find how much time it takes the driver to come to stop (v = 0). We will consider the origin of the reference system as the point at which the driver realizes she must stop. Then x0 = 0

With the equation of velocity, we can obtain the acceleration and replace it in the equation of position, knowing that the position will be 250 m at that time.

v = v0 + a*t

v-v0 / t = a

0 m/s - 71.0 m/s / t =a

-71.0 m/s / t = a

Replacing in the equation for position:

x = v0* t +1/2 * a * t²

250 m = 71.0 m/s * t + 1/2 *(-71.0 m/s / t) * t²

250 m = 71.0 m/s * t - 1/2 * 71.0 m/s * t

250m = 1/2 * 71.0m/s *t

<u>t = 2 * 250 m / 71.0 m/s = 7.04 s</u>

It takes the driver 7.04 s to stop.

Then, we can calculate how much time it took the driver to reach her previous speed. The procedure is the same as before:

v = v0 + a*t

v-v0 / t = a now v0 = 0 and v = 71.0 m/s

(71.0 m/s - 0 m/s) / t = a

71.0 m/s / t =a

Replacing in the position equation:

x = v0* t +1/2 * a * t²

390 m = 0 m/s * t + 1/2 * 71.0 m/s / t * t² (In this case, the initial position is in the pit, then x0 = 0 because it took 390 m from the pit to reach the initial speed).

390m * 2 / 71.0 m/s = t

<u>t = 11.0 s</u>

In total, it took the driver 11.0s + 5.00 s + 7.04 s = 23.0 s to stop and to reach the initial speed again.

In that time, the Mercedes traveled the following distance:

x = v * t = 71.0 m/s * 23.0 s = 1.63 x 10³ m

The Thunderbird traveled in that time 390 m + 250 m = 640 m.

The distance between the two will be then:

<u>distance between both cars = 1.63 x 10³ m - 640 m = 990 m. </u>

3 0

2 years ago