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kobusy [5.1K]
3 years ago
12

A dairy farmer notices that a circular water trough near the barn has become rusty and now has a hole near the base. The hole is

0.38 m below the level of the water that is in the tank. If the top of the trough is open to the atmosphere, what is the speed of the water as it leaves the hole
Physics
1 answer:
ryzh [129]3 years ago
3 0

Answer: 2.73m/s

Explanation:

Potential energy of the water is turned into kinetic energy

mgd = mv^2/2

Where m = mass of water

g = acceleration due to gravity = 9.81m/s^2

d = depth of the water = 0.38m

v = velocity of leaking water

m × 9.81 × 0.38 = mv^2/

eliminate m on both sides and multiply through by 2

7.4456 = v^2

v = sqrt(7.4456)

v = 2.73m/s

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Today, scientists believe that the main reason the Earth has had periodic ice ages is:____________.
horsena [70]

Answer:

Today, scientists believe that the main reason the earth has had periodic ages is that the climate of earth is controlled by the difference in heating of it's surface by the sun. As for example, the equatorial regions are the warmest as the sun is vertically overhead of those areas and the polar regions where the sun is at the extreme angels are the coldest ones.

Explanation:

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carbon dioxide is formed when two oxygen atoms chemically combine with a carbon atom. which term best describes carbon dioxide
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You walk forward at 1.5 m/s for 8s. Your friend decides to walk faster and starts out at 2.0 m/s for the first 4 s. Then she slo
Troyanec [42]
The first thing you should know is that the distance is equal to the speed per time.
 Therefore if You walk forward at 1.5 m / s for 8s
 d = 1.5t [0,8] s
 Your friend decides to walk faster and starts at 2.0 m / s for the first 4 s.
 d = 2t [0,4] s
 Then she slows down and walks forward at 1.0 m / s for the next 4s
 d = t + 4 [4,8] s
 Who walked farther?
 They both walked the same distance
 12 meters

6 0
3 years ago
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7 0
3 years ago
a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^
posledela

Answer:

Approximately 0.8\; \rm s (assuming that air resistance is negligible.)

Explanation:

Let v_0 denote the initial velocity of this ball. Let \theta denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

  • Initial vertical velocity: v_0(\text{vertical}) = v_0 \cdot \sin(\theta).
  • Initial horizontal velocity: v_0(\text{vertical}) = v_0 \cdot \cos(\theta).

If air resistance on this ball is negligible, v_0(\text{vertical}) alone would be sufficient for finding the time of flight of this ball.

Calculate v_0(\text{vertical}) given that v_0 = 8 \; \rm m \cdot s^{-1} and \theta = 30^\circ:

\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}.

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}, the vertical velocity of this ball right before landing would be v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}.

Calculate the change to the vertical velocity of this ball:

\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}.

In other words, the vertical velocity of this ball should have change by 8\; \rm m \cdot s^{-1} during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is g = 10\; \rm m \cdot s^{-2}. In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by 10\; \rm m\cdot s^{-1} each second. What fraction of a second would it take to change the vertical velocity of this ball by 8\; \rm m \cdot s^{-1}?

\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}.

In other words, it would take 0.8\; \rm s to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be 0.8\; \rm s\!.

5 0
3 years ago
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