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melisa1 [442]
2 years ago
5

Two large, plastic tubs were filled with soil The soil was shaped to create a mound in each tub. The starting height of each mou

nd was recorded. The tubs were set outside, side by side The lid was left off of tub A and a lid was placed on top of tub B Once a day, one ounce of water was poured over the top of each mound for ten days straight At the end of the ten days the mound heights were recorded and compared Is this a valid experimental design to test for wind erosion of soil? А) Yes, because dirt is very heavy B) No, because ten days is not long enough U No, since the type of soil was not identified D Yes, steps were outlined for condition with one variable changed lid on or off​
Physics
1 answer:
shutvik [7]2 years ago
6 0

Answer:

b

Explanation:

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A straight wire lies along the y-axis initially carrying a current of 10 A in the positive y-direction. The current decreases an
Elan Coil [88]

Answer:

Explanation:

The magnetic field due to straight wire is into the square coil.

As the current in straight wire decreases the magnetic flux in the coil decreases . The induced magnetic field is into the coil.The induced current is along +y direction

8 0
4 years ago
The motor on a helicopter turns at an angular speed of 6.2 x 102 revolutions per minute. (a) Express this angular speed in radia
arsen [322]

Answer:

  • 64.93 rad/s
  • 38.956 km

Explanation:

(a)

  \dfrac{6.2\cdot 10^2\,\text{rev}}{60\,\text{s}}\times\dfrac{2\pi\,\text{rad}}{\text{rev}}=\dfrac{62\pi\,\text{rad}}{3\,\text{s}}\approx\boxed{64.93\,\text{rad/s}}

__

(b)

  d=r\theta=(3.0\,\text{m})(2.0\cdot 10^2\,\text{s})\left(\dfrac{62\pi\,\text{rad}}{3\,\text{s}}\right)=124\pi\cdot 10^2\,\text{m}=\boxed{38\,956\,\text{m}}

6 0
3 years ago
A tennis ball weighs 2 kg and is traveling at 10 meters per second. What is the tennis ball’s kinetic energy? Find the exponent
adelina 88 [10]
100 J is the amount of kinetic energy
4 0
4 years ago
A 2-kg rock is thrown vertically upward at a speed of 3.2 m/s from the surface of the moon. If it returns to its starting point
worty [1.4K]

Answer:

1.6 m/s2

Explanation:

Let g_m be the gravitational acceleration of the moon. We know that due to the law of energy conservation, kinetic energy (and speed) of the rock when being thrown upwards from the surface and when it returns to the surface is the same. Given that g_m stays constant, we can conclude that the time it takes to reach its highest point, aka 0 velocity, is the same as the time it takes to fall down from that point to the surface, which is half of the total time, or 4 / 2 = 2 seconds.

So essentially it takes 2s to decelerate from 3.2 m/s to 0. We can use this information to calculate g_m

g_m = \frac{\Delta v}{\Delta t} = \frac{0 - 3.2}{2} = \frac{-3.2}{2} = -1.6 m/s^2

So the gravitational acceleration on the Moon is 1.6 m/s2

8 0
3 years ago
A student throws a water balloon with speed v0 from a height h = 1.82 m at an angle θ = 29° above the horizontal toward a target
3241004551 [841]

Answer:

v_o = 7.76 m/s

Explanation:

As we projected the balloon at speed vo at an angle of 29 degree

so the two component of velocity is given as

v_x = v_ocos29 = 0.875 v_o

v_y = v_o sin29 = 0.485 v_o

now we know that in x direction we have

d = v_x t

7.5 = 0.875 v_o t

v_o t = 8.57

in y direction we have

- 1.82 = (0.485 v_o) t - \frac{1}{2}gt^2

-1.82 = 0.485(8.57) - 4.9 t^2

t = 1.1 s

now we have

v_o = 7.76 m/s

7 0
4 years ago
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