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Pepsi [2]
3 years ago
8

Which of the following are species of wild goats? (Select all that apply.)

Physics
2 answers:
Yuliya22 [10]3 years ago
5 0

Siberian Ibex and Himalaya markor are example of species of wild goats.

<h3> What is species of wild goat?</h3>

The wild goat or common ibex refers wild goat species that lives in mountainous areas, forests, shrublands around Turkey and the Caucasus in the west to Turkmenistan, Afghanistan and Pakistan in the east.

They can be found in mountainous area around Africa, Asia, North America.

There are 36 Subfamily of wild goats and they have surefoot which help them to live in mountainous areas.

Therefore, Siberian Ibex and Himalaya markor are example of species of wild goats.

For more details on Wild goats, check the link below.

https://brainly.in/question/6263222

Tomtit [17]3 years ago
4 0

Answer:

Siberian ibex  and Himalayan markhor

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Three liquids are at temperatures of 6 ◦C, 23◦C, and 38◦C, respectively. Equal masses of the first two liquids are mixed, and th
kupik [55]

The equilibrium temperature is T13=3.12 ◦C

<u>Explanation:</u>

<u>Given </u>

The temperature of liquids: T1=6◦C, T2=23◦C, T3=38◦C

The temperature of 1+2 liquids mix: T12= 13◦C.

The temperature of 2+3 liquids mix: T23=26.8 ◦C.

The temperature of 1+3 liquids mix: T13= ??

<u>1.When the first two liquids are mixed:</u>

  • mC1(T1-T12)+mC2(T2-T12)=0
  • C1(6-13)=C2(23-13)=0
  • 7C1=10C2
  • C1=1.42C2

<u>2.When the second and third liquids are mixed</u><u>:</u>

  • mC2(T2-T23)+mC3(T3-T23)=0
  • C2(23-26.8)=C3(38-26.8)=0
  • 3.8C2=12.8C3
  • C2=3.36C3

<u>3.When the first and third liquids are mixed:</u>

  • mC1(T1-T13)+mC3(T3-T13)=0
  • C1(6-T13)+C3(38-T13)=0
  • C1=1.42C2  C2=3.36C3
  • C1=1.42C2(3.36C3)
  • C1=4.77C3
  • C1(6-T13)+C3(38-T13)=0
  • 4.77C3(6-T13)+C3(38-T13)=0
  • By solving the equation we get,
  • T13=3.12 ◦C
  • The equilibrium temperature is T13=3.12 ◦C

<u></u>

7 0
3 years ago
A biology student views several different prokaryotic and eukaryotic cells under a microscope. How do the prokaryotic cells appe
Fofino [41]

Answer:

The prokaryotic cells are smaller than the eukaryotic cells.

Explanation:

Its just the simple process of elimination. Prokaryotic cells lack of a lot of things, including mitochondria, so it couldn't be option C. BOTH prokaryotic and eukaryotic cells have cytosol AND ribosomes, so the answer couldn't be D or A, and you're left with option B. Meaning option B is the correct choice.

7 0
4 years ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
valentina_108 [34]

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

4 0
4 years ago
Define specific heat capacity
sleet_krkn [62]

Answer:you can just look this up yknow?

Explanation:

7 0
3 years ago
Read 2 more answers
What is the heat needed to raise the temperature of 24.7 kg silver from 14.0 degrees celsius to 30.0 degrees celsius? specific h
Citrus2011 [14]
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q=m C_s \Delta T
where
m is the mass of the substance
C_s is its specific heat capacity
\Delta T is the increase of temperature

The sample of silver of our problem has a mass of m=24.7 kg. Its specific heat capacity is C_s = 236 J/g^{\circ}C and the increase in temperature is
\Delta T=30.0^{\circ}-14.0^{\circ}C=16.0^{\circ}C

Therefore, the amount of heat needed is
Q=mC_s \Delta T=(24.7 kg)(236 J/g^{\circ}C)(16.0^{\circ}C)=9.32 \cdot 10^4 J
8 0
4 years ago
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