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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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Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis
solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

v^{2}=u^2+2as

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equalsT=2\times 30.51\approx61seconds

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals 39.66m/s

4 0
3 years ago
Just like energy is <br> matter is lost through an ecosystem
Novosadov [1.4K]
Yes energy is reduced 
7 0
3 years ago
Read 2 more answers
A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera
Fantom [35]
In order to accelerate the dragster at a speed v_f = 100 m/s, its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
W= K_f - K_i = K_f =  \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is -W_f:
W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W

And since 1 horsepower is equal to 746 W, we can rewrite the power as
P=4.08 \cdot 10^6 W \cdot  \frac{1 hp}{746 W} =547 hp



3 0
3 years ago
Read 2 more answers
A rare and valuable antique chest is being moved into a truck using a 4.00 m long ramp. the kj weight of the chest plus packing
Lisa [10]

First let us calculate for the angle of inclination using the sin function,

sin θ = 1 m / 4 m

θ = 14.48°

 

Then we calculate the work done by the movers using the formula:

W = Fnet * d

 

So we must calculate for the value of Fnet first. Fnet is force due to weight minus the frictional force.

Fnet = m g sinθ – μ m g cosθ

Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48

Fnet = 84.526 N

 

So the work exerted is equal to:

W = 84.526 N * 4 m

<span>W = 338.10 J</span>

7 0
3 years ago
A paperboy rode his bike 3m/s. After being chased by a dog for 8 seconds he was traveling 6m/s. What is his accleration
aev [14]
I believe the answer is 0.375 m/s²
8 0
3 years ago
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