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Marina CMI [18]
2 years ago
14

Two students were climbing stairs at school. Student 1 has a weight of 700 N. Student 2 has a weight of 650 N. How much power wo

uld each student have if each took 6 s to climb a 4 m tall flight of stairs?
Student 1 would have 4200 W, and Student 2 would have 3900 W of power.

Student 1 would have 116 W, and Student 2 would have 108 W of power.

Student 1 would have 2800 W, and Student 2 would have 2600 W of power.

Student 1 would have 467 W, and Student 2 would have 433 W of power.
Physics
1 answer:
KIM [24]2 years ago
3 0

Student 1 would have a power 467 W and student 2 would have a power of 433 W. The correct option is the fourth option - Student 1 would have 467 W, and Student 2 would have 433 W of power.

From the question,

We are to calculate the power each student would have to climb the flight of stairs.

Power can be calculated using the formula

P = \frac{F \times d}{t}

Where

P is Power

F is the force

d is the distance

and t is the time

NOTE: The weight of the students represent the force

  • For student 1

F = 700 N

d = 4 m

t = 6 s

∴ P = \frac{700 \times 4}{6}

P = 467 W

  • For student 2

F = 650 N

d = 4 m

t = 6 s

∴ P = \frac{650 \times 4}{6}

P = 433 W

Hence, Student 1 would have a power 467 W and student 2 would have a power of 433 W. The correct option is the fourth option - Student 1 would have 467 W, and Student 2 would have 433 W of power.

Learn more here: brainly.com/question/18801566

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1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

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V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

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