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olga2289 [7]
3 years ago
12

2. Which animal is a primary consumer in the Ethiopian Highlands?

Physics
2 answers:
Morgarella [4.7K]3 years ago
7 0

Answer:

the walia ibex

Explanation:

i failed a quiz it sayed i hope it helps;)

loris [4]3 years ago
6 0

Answer: Wallia Ibex

Explanation: Black lions and tigers are not primary consumers. However, the Wallia Ibex is a primary consumer, hope it helps :D.

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The temperature of water in a beaker is 45°C. What does this measurement represent?
Tasya [4]

Answer:

the average kinetic energy of water particles - C.

8 0
3 years ago
Which planet has the slowest rotation rate?
aliya0001 [1]

Venus has the slowest rotation rate.
It takes Venus  243.7 Earth days to rotate once.

Planet           Period

Venus          243.7 (Earth) days
Mercury        58.8 days
Pluto              6.4 days
Mars            24.6 hours
Earth           23.9 hours
Uranus        17.2 hours
Neptune       16.1 hours
Saturn          10.7 hours
Jupiter            9.9 hours

Interesting . . . the largest planet rotates the fastest.
I wonder why that is.
 
3 0
3 years ago
Read 2 more answers
When not in a vacuum, what force causes objects in freefall to fall at different speeds??
Deffense [45]
Hi,

My best answer would be Gravity. Is it a multiple choice question? Or is it an essay question. 
4 0
3 years ago
A 40kg-skier starts at the top of a 12-meter high slope. At the bottom, she is traveling 10 m/s. How much energy does she lose t
KIM [24]
Energy at top = U = mgh = 40 * 9.8 * 12 = 4704 J

Energy at bottom = 1/2 mv² = 1/2 * 40 * 10² = 4000 / 2 = 2000 J

Energy Lost = Final - Initial = 4704 - 2000 = 2704 J

In short, Your Answer would be Option D

Hope this helps!
4 0
4 years ago
Read 2 more answers
An undersea research chamber is spherical with an external diameter of 5.50 m . The mass of the chamber, when occupied, is 87600
arsen [322]

Answer:

A)

8.75\times10^{5} N

B)

1.7\times10^{4} N

Explanation:

d = diameter of the spherical chamber = 5.50 m

Volume of the spherical chamber is given as

V = \frac{4\pi (\frac{d}{2} )^{3}}{3}\\ V = \frac{4(3.14) (\frac{5.50}{2} )^{3}}{3}\\V = 87.07 m^{3}

\rho = density of the seawater = 1025 kg m⁻³

Part A)

Force of buoyancy by seawater on the chamber is given as

F_{b} = \rho V g \\F_{b} = (1025) (87.07) (9.8)\\F_{b} = 8.75\times10^{5} N

Part B)

T = tension force in the cable in upward direction

m = mass of the chamber = 87600 kg

Weight of the chamber is given as

F_{g} = mg \\

Using equilibrium of force in the vertical direction, we have

T + F_{g} = F_{b}\\T + m g = F_{b}\\T + (87600) (9.8) = 8.75\times10^{5}\\T = 16520 N \\T = 1.7\times10^{4} N

3 0
3 years ago
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