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larisa86 [58]
3 years ago
14

How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of ele

ctrons from an initial point where the electric potential is 6.70 V to a point where the electric potential is -8.90 V? (The potential in each case is measured relative to a common reference point.)
Physics
1 answer:
luda_lava [24]3 years ago
7 0

Answer:

W = 1.5 x 10⁶ J = 1.5 MJ

Explanation:

First, we calculate the potential difference between the given 2 points. So, we have:

V₁ = Electric Potential at Initial Position = 6.7 V

V₂ = Electric Potential at Final Position = - 8.9 V

Therefore,

ΔV = Potential Difference = V₂ - V₁ = -8.9 V - 6.7 V = - 15.6 V

Since, we use magnitude in calculation only. Therefore,

ΔV = 15.6 V

Now, we calculate total charge:

Total Charge = q = (No. of Electrons)(Charge on 1 Electron)

where,

No. of Electrons = Avagadro's No. = 6.022 x 10²³

Charge on 1 electron = 1.6 x 10⁻¹⁹ C

Therefore,

q = (6.022 x 10²³)(1.6 x 10⁻¹⁹ C)

q = 96352 C

Now, from the definition of potential difference, we know that it is equal to the worked done on a unit charge moving it between the two points of different potentials:

ΔV = W/q

W = (ΔV )(q)

where,

W = work done = ?

W = (15.6 V)(96352 C)

<u>W = 1.5 x 10⁶ J = 1.5 MJ</u>

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In the equation for centripetal force, which expression represents the centripetal acceleration of the object? mv2 StartFraction
Sphinxa [80]

Answer: \frac{V^{2}}{r}

Explanation:

According to Newton's 2nd Law of motion the force F is proportional to the mass Fm and acceleration a:

F=m.a (1)

On the other hand, the equation for the Centripetal force is:

F=\frac{mV^{2}}{r} (2)

Where:

V is the velocity

r is the radius of the circular motion

Making (1) and (2) equal:

m.a=\frac{mV^{2}}{r} (3)

Hence:

a=\frac{V^{2}}{r} This is the expression for the centripetal acceleration

It should be noted, this acceleration is directed toward the center of the circumference of the circular motion (that's why it's called centripetal acceleration).

3 0
3 years ago
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The element copper has 29 protons. What is the atomic number of copper?
photoshop1234 [79]
The number of protons an element has is its atomic number so the atomic number would be 29
8 0
3 years ago
A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.
gtnhenbr [62]

Answer:

(a)  vmax = 0.34m/s

(b)  v = 0.13m/s

(c)  v = 0.31m/s

(d)  x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

v_{max}=\omega A       (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

\omega=\sqrt{\frac{k}{m}}           (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

x=Acos(\omega t)

You solve the previous equation for t:

t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s

With this value of t, you can calculate the speed of the object with the following formula:

v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s

The position will be:

x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0
3 years ago
How do I solve this arithmetic sequence?
Serjik [45]

Answer:

The 16ᵗʰ term of this sequence is 82

Step-by-step explanation:

Here,

First Term = a₁ = 9

Common Difference = (d) = 2

Now, For 16ᵗʰ term, n = 16

<em>aₙ = a + (n - 1)d</em>

a₁₆ = 7 + (16 - 1) × 2

a₁₆ = 7 + 15 × 5

a₁₆ = 7 + 75

a₁₆ = 82

Thus, The 16ᵗʰ term of this sequence is 82

<u>-TheUnknownScientist</u>

4 0
2 years ago
Robin would like to shoot an orange in a tree with his bow and arrow. The orange is hanging yf=5.00 myf=5.00 m above the ground.
tensa zangetsu [6.8K]

Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

<u>h' = 55.3 m</u>

4 0
3 years ago
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