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Lunna [17]
3 years ago
5

A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53 degrees above the horizontal. The f

irefighters want to direct the water at a blaze that is 10.0 m above ground level. How far from the building should they position their cannon? There are two possibilities (d1
Physics
1 answer:
Fed [463]3 years ago
3 0

Answer:

8.8 m or 52.5 m from the building

Explanation:

The water cannon shoots at 25m/s at a fixed angle 53 degrees above the horizontal;  the velocity of the water will have vertical components and horizontal component. The vertical component his responsible for the vertical motion of the water.

vertical components of U = U sin 53oc = 25 sin 53oc = 0.7986 × 25 = 19 .96 m/s where U is the initial velocity of the water

using equation of motion and S = height = 10 m

S = ut + 1/2 at²  a = g = -9.81 m/s^2 since it act downward

10 = 19.97t + 1/2 × -9.81 t² = 19.95t - 4.9t²

bring 10 in to form a quadratic equation and multiply equation by (-1)

4.9t² - 19.97t + 10 = 0

solve using quadratic formula

-b±√(b²- 4ac)/2a where a = 4.9 b = 19.97 and c = -10

-19.97±√(398.8 - 196) / ( 2×4.9)

-19.97±√(202.8)/ 9.8

(-19.97±14.24) ÷ 9.8

-34.21/9.8 or -5.73/9.8

t= - 3.49 or t = -0.585s

solving for the horizontal motion

horizontal distance = horizontal component of the velocity × t since it will travel both at the same time

horizontal distance = 25 cos 53oc × 0.585 = 8.8m from the building or

horizontal distance = 25 cos 53oc × 3.49 = 52.5m from the building

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Whitepunk [10]

Explanation:

The point is that water is moving smoothly but that the solutes are not.Even though the containers are chemically different (chemical disequilibrium), once all the solutes in one container are contrasted to all the solutes in another container, both have the same total solutes concentrations (this means that they are in osmotic balance).

8 0
3 years ago
Which would you expect to freeze faster pure water or salt solution why?
krek1111 [17]
Pure water.

A salt solution contains impurities whereas pure water will not contain any impurities.

Impurities increase the boiling point (freezing point) of a substance.

Thus, I would expect the pure water solution to freeze faster than the salt solution.
6 0
3 years ago
When a golf ball is hit, it travels at 41 meters per second. The mass of a golf ball is 0.045kg. Calculate the kinetic energy of
Fittoniya [83]

Answer:

  75.645 J

Explanation:

The kinetic energy is related to the mass and velocity by the formula ...

  KE = 1/2mv²

For the given mass of 0.045 kg, and velocity of 41 m/s, the kinetic energy is ...

  KE = 1/2(0.045 kg)(41 m/s)² = 75.645 J

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The unit of energy, joule, is a derived unit equal to 1 kg·m²/s².

4 0
3 years ago
Adhesion is when forces of the molecules are more strongly drawn to each other than the molecules in other substances true or fa
SSSSS [86.1K]

Answer:

False

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6 0
4 years ago
A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force (F_{B}), as follows:

F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB (1)

<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

5 0
3 years ago
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