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GenaCL600 [577]
3 years ago
5

If I = 2.0 A in the circuit segment shown below, what is the potential difference VB - VA?

Physics
1 answer:
Tom [10]3 years ago
7 0

Answer:

10 V

Explanation:

The potential difference between two points is the amount of work required to carry a unit charge from one point to the other point. This would result in a potential difference between this two points.

The difference between the potential across two points B and A is V_{BA}=V_B-V_A

From the image attached:

V_B-V_A=10-10I+20\\\\But\ I = 2A,hence:\\\\V_B-V_A=10-10(2)+20\\\\V_B-V_A=10-20+20\\\\V_B-V_A=10\ V

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The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed of 7,750 m/s, how long is HS
deff fn [24]

Answer: 5,640 s (94 minutes)

Explanation:

the tangential speed of the HST is given by

v=\frac{2\pi r}{T} (1)

where

2\pi r is the length of the orbit

r is the radius of the orbit

T is the orbital period

In our problem, we know the tangential speed: v=7,750 m/s. The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

r=6.38\cdot 10^6 m+569,000 m=6.95\cdot 10^6 m

So, we can re-arrange equation (1) to find the orbital period:

T=\frac{2 \pi r}{v}=\frac{2 \pi (6.95\cdot 10^6 m/s)}{7,750 m/s}=5,640 s

Dividing by 60, we get that this time corresponds to 94 minutes.

6 0
3 years ago
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A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0
viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
KE = \frac{1}{2}mv^2 \\\\PE = mgh

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

E_f = \frac{1}{2}mv_f^2

And:
W_A = E_i - E_f

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) -  \frac{1}{2}mv_f^2

Solving for the work done by air resistance:
W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) -  \frac{1}{2}(.450)(19.89^2)

W_A = \boxed{42.552 J}

8 0
2 years ago
The radioactive 60co isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequenc
Ivanshal [37]

1. Frequency: 3.23\cdot 10^{20} Hz

The energy given is the energy per mole of particles:

E=1.29\cdot 10^{11} J/mol

1 mole contains a number of Avogadro of particles, N_A, equal to

N_A=6.022\cdot 10^{23} particles

So, by setting the following proportion, we can calculate the energy of a single photon:

1.29 \cdot 10^{11} J/mol : 6.022 \cdot 10^{23} ph/mol = E_1 : 1 ph\\E_1 = \frac{(1.29\cdot 10^{11} J/mol)(1 ph)}{6.022\cdot 10^{23} ph/mol}=2.14\cdot 10^{-13} J

This is the energy of a single photon; now we can calculate its frequency by using the formula:

E_1 = hf

where

h=6.63\cdot 10^{-34} Js is the Planck's constant

f is the photon frequency

Solving for f, we find

f=\frac{E_1}{h}=\frac{2.14\cdot 10^{-13} J}{6.63\cdot 10^{-34} Js}=3.23\cdot 10^{20} Hz

2. Wavelength: 9.29\cdot 10^{-13} m

The wavelength of the photon is given by the equation:

\lambda=\frac{c}{f}

where

c=3\cdot 10^8 m/s

is the speed of the photon (the speed of light). Substituting,

\lambda=\frac{3 \cdot 10^8 m/s}{3.23\cdot 10^{20} Hz}=9.29\cdot 10^{-13} m

6 0
3 years ago
An Ohmic material is one in which its resistance does not depend on the voltage across it or the current within it. In that case
Alexandra [31]

Voltage

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Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
3 years ago
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