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Lina20 [59]
3 years ago
10

Imagine that you are in a bleachers watching a swim meet in which your friend is competing in the freestyle event. At the instan

t the starting gun fires, the lights go out! When the lights come back on, the timer on the scoreboard records 86s. You observe your friend is now about halfway along the length of pool, swimming in a direction opposite to that in which he started. The pool is 50 m long.
what are two possible distances that you might infer your friends swam while the lights were out, given the average velocity is 0.29 meters per second
Physics
1 answer:
kramer3 years ago
7 0

The two possible distances that you might infer your friends swam while the lights were out are 25 m and 75 m.

Answer:

Explanation:

So you have to measure the distance covered by your friend in a time gap of 86 s. And the average velocity is given as 0.29 m/s.

Then as per the mathematical calculation of velocity, distance can be measured as the product of velocity with time interval.

Distance = Velocity × Time Interval

Distance = 0.29 m/s×86 s = 24.94 m.

So based on this calculation, one of the possible distance inferred by you will be 24.94 m.

Another possible distance can be guessed from the statements provided. So if the length of pool is 50 m, then covering halfway in opposite direction to his starting direction means completion of one full length i.e., 50 m and then halfway of that 50 m which is 25m, so totally 50 +25 = 75 m.

So in other way, we can assume that your friend has covered 75 m distance during the light out.

Thus, the two possible distances that you might infer your friends swam while the lights were out are 25 m and 75 m.

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A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters​ [cm]. The rover picks up the rock and lifts it
Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid SG (also called relative density) is the ratio of the density of that solid \rho_{rock} to the density of water \rho_{water}=1 kg/m^{3} (normally at 4\°C):

SG=\frac{\rho_{rock}}{\rho_{water}} (1)

On the other hand, the density of the rock is calculated by:

\rho_{rock}=\frac{m_{rock}}{V_{rock}} (2)

Where:

m_{rock} is the mass of the rock

V_{rock}=\frac{4}{3} \pi r^{3} is the volume of the rock, since is spherical

Well, we already know the value of \rho_{water}, but we need to find \rho_{rock} in order to find the rock's specific gravity; and in order to do this, we firsly have to find m_{rock} and then calculate V_{rock}:

In the case of the mass of the rock, we can calclate it by the following equation:

W_{rock}=m_{rock}g_{mars} (3)

Where:

W_{rock} is the weight if the rock in mars

g_{mars}=3.7 m/s^{2} is the acceleration due gravity in Mars

Isolating m_{rock}:

m_{rock}=\frac{W_{rock}}{g_{mars}} (4)

m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}} (5)

To find W_{rock} we can use the following equation of the potential gravitational energy U:

U=W_{rock}H (6)

Where:

U=2 J=2 Nm is the potential energy

H=20 cm \frac{1m}{100 cm}=0.2 m is the height at which the rock has the mentioned potential energy

Isolating W_{rock}:

W_{rock}=\frac{U}{H} (7)

W_{rock}=\frac{2 Nm}{0.2 m} (8)

W_{rock}=10 N (9)

Substituting (9) in (5):

m_{rock}=\frac{10 N}{3.7 m/s^{2}} (10)

m_{rock}=2.702 kg (11)

Substituting (11) in (2):

\rho_{rock}=\frac{2.702 kg}{V_{rock}} (12) At this point we only need to find the volume of the rock, knowing its diameter is d=10 cm, hence its radius is r=\frac{d}{2}=5 cm

V_{rock}=\frac{4}{3} \pi (5 cm)^{3} (13)

V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3} (14)

Substituting (14) in (12):

\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}} (15)

\rho_{rock}=5166.34 kg/m^{3} (16)

Substituting (16) in (1):

SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}} (17)

Finally we obtain the specific gravity of the​ rock:

SG=5166.347

7 0
3 years ago
Vocabulary word for the total amount of charge in a closed system remains constant
Usimov [2.4K]

Answer:

i think this is it i dont know tho A conservation law stating that the total electric charge of a closed system remains constant over time, regardless of other possible changes within the system. "Conservation of charge." YourDictionary. LoveToKnow

7 0
3 years ago
How will the behavior of the gases in the piston chamber change if more pressure is applied to the piston?
allsm [11]
Here molecules of gas remains constant. If more pressure is applied to the piston, gaseous molecules will come closer to each other. Hence their volume decreases and density increases. Hope this helps you.
6 0
3 years ago
Read 2 more answers
Global positioning satellites (GPS) can be used to determine your position with great accuracy. If one of the satellites is 20,0
o-na [289]

Answer:

0.00001 %

Explanation:

The distance from satellite = 20000 km

Position range = 2 m

The percentage uncertainty is given by dividing the distance from satellite by the position range of desired accuracy.

Percentage uncertainty is given by

\dfrac{2}{20000\times 10^3}\times 100=0.00001\ \%

The percentage uncertainty that is required is 0.00001 %

8 0
4 years ago
Vector a with arrow has a magnitude of 12.3 units and points due west. vector b with arrow points due north. (a) what is the mag
lesantik [10]

part a)

Vector a has magnitude 12.3 and its direction is west, while Vector b has unknown magnitude and its direction is north. This means that the two vectors form a right-angle triangle, so a and b are two sides, while a+b is the hypothenuse.

We know the magnitude of a+b, which is 14.5, so we can use the Pythagorean theorem to calculate the magnitude of b:

|b|=\sqrt{(a+b)^2-a^2}=\sqrt{(14.5)^2-(12.3)^2}=7.68


part b) The direction of the vector a+b relative to west can be found by calculating the tangent of the angle of the right-angle triangle described in the previous part; the tangent of the angle is equal to the ratio between the opposite side (b) and the adjacent side (a):

tan x=\frac{b}{a}=\frac{7.68}{12.3}=0.62

And the angle is

x=tan^{-1} (0.62)=31.8^{\circ}

with direction north-west.


part c)

This is exactly the same problem as the one we solved in part a): the only difference here is that the hypothenuse of the triangle is now given by a-b rather than a+b. In order to find a-b, we have to reverse the direction of b, which now points south. However, the calculations to get the magnitude of b are exactly the same as before, since the magnitude of (a-b) is the same as (a+b) (14.5 units), therefore the magnitude of b is still 7.68 units.


part d)

Again, this part is equivalent to part b); the only difference is that b points now south instead of north, so the vector (a-b) has direction south-west instead of north-west as before. Since the magnitude of the vectors involved are the same as part b), we still get the same angle, 31.8^{\circ}, but this time the direction is south-west instead of north-west.

5 0
4 years ago
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