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bogdanovich [222]
3 years ago
12

Suppose Joe, who can type 40 words per minute, takes 2.5 hours to type his report. To speed things up, his friend Mark offers to

help him. Mark can type 20 words per minute.Working together, how long will it take Joe and Mark to type this report?
Physics
1 answer:
Vika [28.1K]3 years ago
5 0

Explanation:

When Joe works alone, the total number  of words he typed can be given by:

Total words = (40 words per minute) x (60 minutes per hour) x (2.5 hours)

Total words = 6000 words

Now, when Joe and Mark work together, let 'y' be the number of hours for which they both work simultaneously:

Total words = Words Typed by Joe + Words Typed by Mark

6000 = {(40 words per minute) x (60 minutes per hours) x (y hours)} + {(20 words per minute) x (60 minutes per hours) x (y hours)}

6000 = 2400y + 1200y = 3600y

y = 1.67 hours = 1 hour and 40 minutes

Thus, working together simultaneously, Joe and Mark will take 1 hour and 40 minutes to complete the report.

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A spring with a mass on the end of it hangs in equilibrium a distance of 0.4200 m above the floor. The mass is pulled down a dis
den301095 [7]

Answer:

0.48 m

Explanation:

I'm assuming that this takes place in an ideal situation, where we neglect a host of factors such as friction, weight of the spring and others

If the mass is hanging from equilibrium at 0.42 m above the floor, from the question, and it is then pulled 0.06 m below that particular position. This pulling is a means of adding more energy into the spring, when it is released, the weight compresses the spring and equals its distance (i.e, 0.06 m) above the height.

0.42 m + 0.06 m = 0.48 m

At the highest point thus, the height is 0.48 m above the ground.

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3 years ago
Four point charges of magnitudes +3q, -q, +2q, and -4q are arranged in the corners of a square of side length L. The charge -q c
mafiozo [28]

Answer:

d) 0 V

Explanation:

It can be showed that the potential due to a point charge q, to a distance d from the charge, can be expressed as follows:

V = \frac{k*q}{r}

where k = \frac{1}{4*\pi*\epsilon0} = 9e9 N*m2/C2

As the potential is an scalar, and is linear with the charge, we can apply the superposition principle, which means that we can find the potential due to one of the charges, as if the other were not present.

By symmetry, all four charges are at the same distance from the center, so we can write the total potential, as follows:

V = \frac{k}{d} ( q1 + q2 + q3 + q4) (1)

where d, is the semi-diagonal of the square, that we can find applying Pythagorean theorem, as follows:

d = \sqrt{\frac{L^{2}}{4} + \frac{L^{2}}{4} } = L*\frac{\sqrt{2}}{2}

Replacing by the values in (1) we have:

V = \frac{9e9N*m2/C2}{\frac{L}{2}*\sqrt{2} }* ( +3q -q + 2q + -4q)  = 0 V

which is equal to the option d).

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Answer:

The answer is B, although technically that is an eclipse.

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A charge of 1.0 × 10-6 μC is located inside a sphere, 1.25 cm from its center. What is the electric flux through the sphere due
Korvikt [17]

Answer:

\Phi_E=0.11\frac{N\cdot m^2}{C}

Explanation:

According to Gauss's Law, the electric flux of a charged sphere is the electric field multiplied by the area of ​​the spherical surface:

\Phi_E=EA\\\Phi_E=E(4\pi R^2)\\\Phi_E=\frac{q}{4\pi \epsilon_0 R^2}(4\pi R^2)\\\Phi_E=\frac{q}{\epsilon_0}

This is identical to the electric flux of a point charge located in the center of the sphere.

\Phi_E=\frac{1*10^{-12}C}{8.85*10^{-12} \frac{C^2}{N\cdot m^2}}\\\Phi_E=0.11\frac{N\cdot m^2}{C}

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