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xxMikexx [17]
3 years ago
11

Consider a river flowing toward a lake at an average velocity of 3 m/s. the river height is 90m above the lake. what is the tota

l mechanical energy of the river water per unit mass in kj/kg?
Physics
1 answer:
strojnjashka [21]3 years ago
6 0

Kinetic energy per unit of mass is

K=\frac{v^{2} }{2}

Given, v=3m/s^{2}

Therefore,

K=\frac{(3^{2} m/s^{2} )^2}{2}

K=4.5 J/kg

Now potential energy per unit mass is

p=g\times h

Given, h=90 m

Therefore,

p= 9.8m/s^2 \times 90

p=882.9 J/kg

Thus, total mechanical energy of the river water per unit mass is

T=K+p=(4.5+882.9)J/kg

T=887.9 J/kg

OR

T=0.887 kJ/kg

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7 0
3 years ago
Imagine Two Artificial Satellites Orbiting Earth At The Same Distance. One Satellite Has A Greater Mass Than The Other One? Whic
Zarrin [17]
1. B
It comes from Newton's law of universal gravitation.
<span><span>,
</span></span>F=G\frac{m_{1}m_{2}}{r^2}<span><span>
</span></span>
Where:
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2. D.
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3 0
4 years ago
Calculate the volume of this regular solid.
fenix001 [56]

Answer:

2122.64 cm^3

Explanation:

Some data in the problem are missing.

Missing values:

Radius of the cylinder: 13 cm

Height of the cylinder: 4 cm

The volume of a cylinder is given by

V=\pi r^2 h

where

r is the radius of the base of the cylinder

h is the height of the cylinder

In this problem, we have:

r = 13 cm (radius)

h = 4 cm (height)

Therefore, the volume of the cylinder is:

V=\pi (13)^2(4)=2122.64 cm^3

6 0
3 years ago
A rocket accelerates upward from rest at a rate of 5 m/s^2. At a height of 1000m, the engines burn out and it coasts the rest of
laila [671]

Answer:

Explanation:

Given

acceleration of rocket(a)=5 m/s^2

At h=1000 m rocket burn out

v^2-u^2=2as

v^2-0=2\times 5\times 1000

v=\sqrt{10^4}=100 m/s

(b) time to reach v=100 m/s

v=u+at

100=0+5\times t

t=\frac{100}{5}=20 s

(c)Rocket maximum altitude

v^2-u^2=2as

here u=100 m/s

v=0

a=9.8 m/s^2

s=\frac{v^2}{2g}

s=\frac{100^2}{2\times 9.8}

[tex]s=510.204

Therefore maximum altitude=510.204+1000=1510.204 m

7 0
3 years ago
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