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-Dominant- [34]
3 years ago
6

A trough is 10 feet long and its ends have the shape of isosceles triangles that are 3 feet across at the top and 1 foot high. I

f the trough is filled with water at a rate of 12 feet cubed per minute, how fast is the water level rising when the trough is half a foot deep?
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
6 0

Answer:

1.6 ft/min

Explanation:

Since trough is 10 ft long and water is filled at the rate of 12ft3/min. We can calculate the rate of water filled with respect to area:

= 12 / 10 = 1.2ft2/min

As the water level rises, so does the water surface, or the bottom side of the isosceles triangles. In fact we can calculate the bottom side when the trough is half foot deep:

= 3 / 2 = 1.5 ft

The rate of change in water level would be the same as calculating the height of the isosceles triangles knowing its base

= 1.2 * 2 / 1.5 = 1.6 ft/min

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Consider a loop of wire whose plane is horizontal and that carries a current in the clockwise direction when viewed from above.
Mars2501 [29]

Answer:

Downwards into the plane

Explanation:

Solution:-

- This is a conceptual application of  hand rule. We will place our palm fingers open vertical to a plane surface. Then curl our fingers in and naturally point the thumb.

- The direction of curl of fingers denotes the direction of of current flow in the coil. Which in our case is "clockwise direction". We will orient/invert our right hand palm in such a way that we curl our fingers in clockwise fashion. Then stick the thumb out to give us the direction of magnetic field or North pole end. In our case the the thumb points downwards into the plane denoting that the magnetic field within the loop is also acting downwards into the plane.

- The bar magnet would be placed in such a way that North pole is pointing downward into the plane in the direction of magnetic field and end up at south pole pointing up out of the plane.

7 0
3 years ago
Gayle cooks in her micwave
liubo4ka [24]
The correct spelling is "Microwave".
4 0
3 years ago
An attacker at the base of a castle wall 3.95 m high throws a rock straight up with speed 5.00 m/s from a height of 1.60 m above
s344n2d4d5 [400]

Answer:

a) No

b) the rock must have a minimum initial speed of 6.79m/s for it to reach the top of the building.

Explanation:

Given:

Height of the wall = 3.95m

Initial height = 1.60m

Initial speed = 5.00m/s

distance between the initial height and wall top = 3.95 - 1.60 = 2.35m

Using the formula;

v^2 = u^2 + 2as ....1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance travelled

From equation 1

s = (v^2 - u^2)/2a ...2

Since the rock t moving up,

the acceleration = -g = -9.8m/s2

s = maximum height travelled

v = 0 (at maximum height velocity is zero)

Substituting into equation 2

s = (0 - 5^2)/(2×-9.8) = 1.28m

Therefore, the maximum height is 1.28 from his initial height Which is less than the 2.35m of the wall from his initial height. So the rock will not reach the top of the wall

b) Using equation 1:

u^2 = v^2 - 2as

v = 0

a = -9.8m/s

s = 2.35m. (distance between the initial height and wall top)

u^2 = 0 - 2(-9.8 × 2.35)

u^2 = 46.06

u = √46.06

u = 6.79m/s

Therefore, the rock must have a minimum initial speed of 6.79m/s

8 0
3 years ago
A car's acceleration is 3m/s2. If the car started at rest and it only took 10s for the car to reach this acceleration, what is t
Grace [21]

Answer:

30m/s

Explanation:

From law of motion equation

Vf= Vi + at

Where Vf= final velocity

Vi= initial velocity=0(the car started at rest)

a= acceleration= 3m/s2

t= time= 10s

Then substitute into the equation to get the final velocity.

Vf= 0+(10×3)

Vf= 30m/s

Hence, the car's final velocity is 30m/s

5 0
3 years ago
A dog and a cat sit on a merry-go-round. The dog sits 0.3 m from the center. The cat sits 1.5 m from the center. The whole merry
pochemuha

Answer:

hope that pic helps!! let me know!!

Explanation:

or this expert answers from expert

8 0
3 years ago
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