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Alenkinab [10]
2 years ago
5

A garden hose has a radius of 0.0120 m, and water initially comes out at a speed of 2.88m/s. Dasha puts her thumb over the end ,

which cuts its area to 1.05×10^-4 m^2.what is the new velocity of the water coming out ?
Physics
1 answer:
creativ13 [48]2 years ago
3 0

Answer:

v = 12.4 [m/s]

Explanation:

With the speed and Area information, we can determine the volumetric flow.

V=v*A\\A=\pi *r^{2}

where:

r = radius = 0.0120 [m]

v = 2.88 [m/s]

A=\pi *(0.0120)^{2} \\A=4.523*10^{-4} [m]\\

Therefore the flow is:

V=2.88*4.523*10^{-4} \\V=1.302*10^{-3} [m^{3}/s ]

Despite the fact that you cover the inlet with the finger, the volumetric flow rate is the same.

v=V/A\\v=1.302*10^{-3} /1.05*10^{-4} \\v=12.4[m/s]

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pentagon [3]

Answer:

C.

Explanation:

Option C

The information given in the question tells us about the number of electrons in an atom and also the number of shells in the atom. So, we will come to know about the atomic number, size and chemical properties of the atom. But we cannot determine atomic mass. Atomic mass is a function of number of neutrons and protons.

3 0
2 years ago
A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is
pochemuha

Answer:

2856.96 J

0

0

\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f

6.78822 m/s

Explanation:

v_i = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J

At height y = 0 the potential energy is 0 as

P=mgy\\\Rightarrow P=mg0=0

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f

Half of maximum height

\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}

h_i=\frac{v_i^2}{2g}

v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s

The velocity of the athlete at half the maximum height is 6.78822 m/s

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The clownfish is classified as an omnivore because it eats
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Because it eats plants and animals. 
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A student standing on a cliff that is a vertical height d = 8.0 m above the level ground throws a stone with velocity v0 = 24 m/
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Answer:

a) Vf = 27.13 m/s

b) It would have been the same

Explanation:

On the y-axis:

Y=-Vo*sin\theta*t-1/2*g*t^2

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Solving for t:

t1 = 0.67s     t2= -2.4s

Discarding the negative value and using the positive one to calculate the velocity:

Vf_y = -Vo*sin\theta-g*t

Vf_y = -15.3m/s

So, the module of the velocity will be:

Vf=\sqrt{(-15.3)^2+(24*cos(21))^2}

Vf=27.13m/s

If you throw it above horizontal, it would go up first, and when it reached the initial height, the velocity would be the same at the throwing instant. And starting then, the movement will be the same.

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3 years ago
Which type of wave does not require a medium to travel through
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Electromagnetic waves don't require medium to travel
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