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Ainat [17]
3 years ago
8

What would happen if there were no friction between the girl and the slope?

Physics
1 answer:
erica [24]3 years ago
7 0
She'll most likely fall
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It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing
babunello [35]

Answer:

3 km/h

Explanation:

Let's call the rowing speed in still water x, in km/h.

Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

8/(x - 2) + 8/(x + 2) = 48/5      (notice that: time = distance/speed)

Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)

8*(x+2) + 8*(x-2) =  (48/5)*(x² - 4)

Dividing  by 8

(x+2) + (x-2) = (6/5)*(x² - 4)

2*x = (6/5)*x² - 24/5

0 =  (6/5)*x² - 2*x - 24/5

Using quadratic formula

x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}

x = \frac{2 \pm 5.2}{2.4}

x_1 = \frac{2 + 5.2}{2.4}

x_1 = 3

x_2 = \frac{2 - 5.2}{2.4}

x_2 = -1\; 1/3

A negative result has no sense, therefore the rowing speed in still water was 3 km/h

7 0
3 years ago
Mercury’s natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) c
Georgia [21]

Answer:

Mercury's natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) could mercury naturally exist?

Liquid is the answer

Explanation:

4 0
2 years ago
Consider the circuit shown below V=18.00V. The terminal voltage of the battery is.
Pavel [41]
Resistance is current x potential difference. So therefor run wafff
6 0
2 years ago
Read 2 more answers
How much does the speed of a car increase if it accelerates
Nata [24]

Answer:

12.5 m/s

Explanation:

a = Δv / Δt

2.5 m/s² = Δv / 5 s

Δv = 12.5 m/s

4 0
3 years ago
You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the eart
hichkok12 [17]

Answer:

v = 3.5 \times 10^5 m/s

Explanation:

At some distance from the Earth the force of attraction due to moon is balanced by the force due to Moon

so we will have

\frac{GM_em}{r^2} = \frac{GM_m}{(d-r)^2}

now we have

\frac{d - r}{r} = \sqrt{\frac{M_m}{M_e}}

\frac{3.844\times 10^8 - r}{r} = \sqrt{\frac{7.36 \times 10^{22}}{5.9742\times 10^{24}}}

so we will have

r = 3.46 \times 10^8 m

Now by energy conservation

-\frac{GM_e}{R_e} - \frac{GM_m}{d - (R_e + R_m)} + \frac{1}{2}v^2 = -\frac{GM_e}{r} - \frac{GM_m}{d - r}

-6.26 \times 10^{8} - 13046 + \frac{1}{2}v^2 = -1.15 \times 10^6 - 1.28 \times 10^5

v = 3.5 \times 10^5 m/s

7 0
3 years ago
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