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Vikentia [17]
3 years ago
10

Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4

.50 cm in diameter? (b) At what speed will it shoot out if the di- ameter of the hole is three times as large?
Physics
1 answer:
kati45 [8]3 years ago
5 0

Answer:

velocity  = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = \frac{\pi }{4} × (4.50×10^{-2})²  × velocity

solve it we get

velocity  = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = \frac{\pi }{4} × 3 ×  (4.50×10^{-2})²  × velocity

velocity = 52.4 m/s

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Psychiatry assesses and treats people with mental disorders
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4 years ago
An elevator car weighs 5500 N. If the car accelerates upwards at a rate of 4.0 m/s2, what is the tension in the support cable li
nadya68 [22]

Answer:

Explanation:

The equation for this, since we are talking about weight on an elevator, is Newton's 2nd Law adjusted to fit our needs:

F_n=ma+w where the Normal Force needed to lift that elevator car is the tension. So the equation then becomes

T = ma + w where T is the tension in the cable to lift the elevator, m is the mass of the elevator (which we have to solve for), a is the acceleration of the elevator (positive since it's going up), and w is the weight of the elevator (which we have as 5500 N). Solving first for mass:

w = mg and

5500 =- m(10) so

m = 550 kg. Now we have what we need to solve for the tension:

T = 550(4.0) + 5500 and

T = 2200 + 5500 so

T = 7700 N

4 0
3 years ago
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re
spayn [35]

The magnitude of the electric field can be calculated using the equation

E = \frac{F}{q}, where F is the Force acting on the proton and q is the charge of the proton.

We know that the charge of the proton is q = 1.6 x 10^{-19}  C

We have to calculate the Force first.

We know that F = ma from Newton's 2nd law, where m is the mass of the proton and a is the acceleration.

We know that the mass of the proton m = (1.67) X 10^{-27}  kg

So it turns out that we have to calculate acceleration before anything else.

In order to calculate the acceleration, we make use of the following data from the question:

Initial Velocity of the proton V_{i}  = 0

Distance traveled  D = 1.70 cm = 0.017 m

Time taken for the travel between the plates t = (1.48) X 10^{-6}  s

Acceleration a = ?

Using the equation, D = V_{i}t + \frac{1}{2} at^{2}, we get

Knowing that initial velocity is 0, the equation reduces to D = \frac{1}{2}at^{2}

Rearranging the equation so as to make a the subject of the formula, we have

a = \frac{2D}{t^{2} }

Plugging in the numbers and simplifying gives us a = 1.5 x 10^{10}   m/s^{2}

We can now calculate the Force using F = ma

Plugging in the known values, we get F = 2.5 x 10^{-17}  N

Using this, we can calculate E through the equation E = \frac{F}{q}

Plugging the numbers and simplifying gets us E = 156.25 N/C

Thus, the magnitude of the electric field between the plates of the capacitor is 156.25 N/C


B) To calculate the Final Velocity of the proton, we can make use of the equation

V_{f}  = V_{i}  + at

Plugging the numbers in and simplifying gets us V_{f}  = (2.22)  *  10^{4}  m/s

5 0
4 years ago
Read 2 more answers
Use the equation PiVi PfVf. Assume that Pi 101 kPa and Vi 10.0 L. If Pf 43.0 kPa, what is Vf
iren2701 [21]
Solving for vf gives you PiVi/Pf. Now plug in 101kPa*10L/43kPa = 23.48L. Using significant figures i would round to 23.5L
8 0
4 years ago
A trio of students push a 65 kg crate. The first student pushes 31 N [e], the second student pushes 28 N [s] and the third stude
Verdich [7]

Answer:

The free body diagram is attached.

Explanation:

A force of 31[N] to the east, the second force goes to the south and it is equal to 28[N], the third force goes to the west and it is equal to 39 [N].

We can consider the crate as a particle. And all the forces are acting over the particle.

5 0
3 years ago
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