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sergejj [24]
3 years ago
5

As part of a daring rescue attempt, the Millennium Eagle coasts between a pair of twin asteroids, as shown in the figure below w

here d1 = 2.55 km and d2 = 1.28 km. The mass of the spaceship is 2.53 x10^7 kg and the mass of each asteroid is 3.70x10^11 kg. Find the speed of the Millennium Eagle at point A if its speed at point B is 0.990 m/s.
Physics
1 answer:
Luden [163]3 years ago
4 0
The correct answer is 3 wiv 2=6
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A student walks 4 block east, 7 blocks west, 1 blocks east then 2 blocks west in an hour. Her average speed is____ block/hour
Gemiola [76]

Average speed = (total distance) / (total time)

Average speed = (4+7+1+2 blox) / (1 hour)

<em>Average speed = 14 blocks/hour</em>

<em></em>

I'm gonna go out on a limb here and take a wild guess:

I'm guessing that there's another question glued onto the end of this one, and it asks you to find either her displacement or her average velocity.  I'm so sure of this that I'm gonna give you the solution for that too.  If there's no more question, then you won't need this, and you can just discard it.  I won't mind.

Average velocity = (displacement) / (time for the displacement)

"Displacement" = distance and direction from the start point to the end point, regardless of how she got there.

Displacement = (4E + 7W + 1E + 2W)

Displacement = (5E + 9W)

<em>Displacement = 4 blocks west</em>

Average velocity = (4 blocks west) / (1 hour)

<em>Average velocity = 4 blocks/hour  West</em>

4 0
2 years ago
A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.
zzz [600]

Answer:

0.0257259766982 m

Explanation:

P_2 = Atmospheric pressure = 101325 Pa

d_1 = Initial diameter = 1.5 cm

d_2 = Final diameter

\rho = Density of water = 1000 kg/m³

h = Depth = 40 m

The pressure is

P_1=P_2+\rho gh\\\Rightarrow P_1=101325+1000\times 9.81\times 40\\\Rightarrow P_1=493725\ Pa

From ideal gas law we have

\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow \dfrac{P_1\dfrac{4}{3\times8}\pi d_1^3}{T_1}=\dfrac{P_2\dfrac{4}{3\times8}\pi d_2^3}{T_2}\\\Rightarrow \dfrac{P_1d_1^3}{T_1}=\dfrac{P_2d_2^3}{T_2}\\\Rightarrow d_2=(\dfrac{P_1d_1^3T_2}{P_2T_1})^{\dfrac{1}{3}}\\\Rightarrow d_2=(\dfrac{493725\times 0.015^3\times (20+273.15)}{101325\times (10+273.15)})^{\dfrac{1}{3}}\\\Rightarrow d_2=0.0257259766982\ m

The diameter of the bubble is 0.0257259766982 m

8 0
3 years ago
A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul
GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh

here we know that for spherical shell and pure rolling conditions

v = R \omega

I = \frac{2}{3}mR^2

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh

\frac{5}{6}mv^2 = mgh

h = \frac{5v^2}{6g}

h = \frac{5(8^2)}{6(9.81)} = 5.44 m

Part b)

If ball is not rolling and just sliding over the hill then in that case

\frac{1}{2}mv^2 = mgh

h = \frac{v^2}{2g}

h = \frac{8^2}{2(9.81)} = 3.16 m

3 0
3 years ago
Recording the location of a star requires a measurement of:
Scorpion4ik [409]

Answer:

It requires a measurement of altitude azimuth time.

Hope this helps, if it did, please give it a brainliest.

7 0
2 years ago
I’m designing a kitchen for a person in a wheel chair, which accommodation is least important? A. Provide turn around space for
Valentin [98]
The least important accommodation would be C. 
7 0
3 years ago
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