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vovikov84 [41]
3 years ago
5

A wildebeest runs with an average speed of

Physics
1 answer:
vagabundo [1.1K]3 years ago
4 0
How many seconds did the wildebeest run??
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Which is the truest statement? Standing is health-enhancing because it uses energy. Jogging is not a good way tp enhance your he
Setler [38]
A. <span>Standing is health-enhancing because it uses energy.

Because standing can burn a lot of calories.</span>

5 0
3 years ago
A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled?
zhannawk [14.2K]

Answer:

A boat travels for three hours with a... A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in still water?

Explanation:

5 0
3 years ago
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Which of the following is a theory stating that one plate is forced beneath another plate?
morpeh [17]

Answer:

theroy of plate tectonics

7 0
3 years ago
Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass
Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)

\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)

P = 57.1192W

Therefore the power of the pump is 57.11W

6 0
3 years ago
A perfectly elastic collision occurs between a 15.0-kg mass traveling at 3.50 m/s and a 9.00-kg mass traveling at 2.35 m/s. if t
BaLLatris [955]
Momentum is conserved in a collision. Momentum is mass*velocity, so you can find your answer by calculating initial and final momentums and setting them equal to each other.

15kg * 3.50 m/s + 9kg * 2.35 m/s = 73.65 kg m/s

73.65 = 9 * 2.8 + 15x

solve for x
x= 3.23

The final velocity is 3.23 m/s
5 0
3 years ago
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