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raketka [301]
3 years ago
11

The flow rate in a firehose is 0.524 m3/s. It is able to shoot water to the top of a building 40.4 m tall, but not higher. You r

eplace the nozzle with another one that has a diameter twice as great. What is the highest point the water can reach?
Physics
1 answer:
Marrrta [24]3 years ago
3 0

Answer:

A fire hose must be able to shoot water to the top of a building 35.0 m tall ... Water enters this hose at a steady rate of 0.500 m3/s and shoots out of a round nozzle. ... I know that Flow rate=0.500 m3/s=A*V. I know the pressure needed to ... The first equation has no potential while the second has no kinetic.

Explanation:

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What is the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure
mezya [45]

Answer:

this is a no brainer

Explanation:

As air pressure in an area increases, the density of the gas particles in that area increases.

8 0
3 years ago
A force of 5 n produces an acceleration of 8m/s2 in mass m1 and an acceleration of 24 m/s2 in mass m2 .what acceleration would i
Inessa [10]

Acceleration of the both masses tied together= 6m/s²

Explanation:

The force is given by F= ma

so 5= m1 (8)

m1=0.625 Kg

for m2

5=m2 (24)

m2=0.208 kg

now total mass= m1+m2=0.625+0.208

Total mass=M=0.833 Kg

now F= ma

5= 0.833 (a)

a= 5/0.833

a=6m/s²

4 0
3 years ago
What is meant by resistance and voltmeter?​
LUCKY_DIMON [66]

Answer:

<em>The internal resistance of an ideal ammeter will be zero since it should allow current to pass through it. Voltmeter measures the potential difference, it is connected in parallel. .</em>

Explanation:

<h3><em>I </em><em>hope</em><em> this</em><em> helps</em><em>!</em></h3>
7 0
2 years ago
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

Separation between slits, d=2.2\times 10^{-5}\ m

(a) The condition for maxima is given by :

d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

\theta=3.90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an ini
Volgvan

Answer:

The horizontal range will be 2.55\times 10^5m

Explanation:

We have given initial speed of the shell u = 1.6\times 10^3m/sec

Angle of projection = 51°

Acceleration due to gravity g=9.8m/sec^2

We have to find maximum range

Horizontal range in projectile motion is given by

R=\frac{u^2sin2\Theta }{g}=\frac{(1.60\times 10^3)^2sin(2\times 51^{\circ})}{9.81}=2.55\times 10^5m

So the horizontal range will be 2.55\times 10^5m

6 0
3 years ago
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