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kap26 [50]
3 years ago
15

Owen is going on a multi-day hike up a large mountain. He wants to know what clothes he should bring for the higher elevation as

he hikes. What clothes should Owen bring? (2 points)
a
An umbrella, since precipitation is more common at high elevation.

b
Heavier clothes, since the higher elevation will have cooler temperatures.

c
The same type of clothes he wears normally, as the temperature will remain the same at high elevation.

d
Lighter clothes, because the higher elevation will have warmer temperatures.
Physics
2 answers:
faltersainse [42]3 years ago
8 0
The answer is c hope this helps
zimovet [89]3 years ago
4 0
I think it’s c i’m sorry if it’s not...
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Why does sound travel faster in water than in air?
mars1129 [50]

Answer:

C.

Explanation:

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3 years ago
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On the way to school, Jed traveled 100m north, 300m east, 100 north, 100m east, and 100m north. A.) Find the total distance trav
Oduvanchick [21]

Answer:

Total distance = 700 m

Displacement = 500 m

Explanation:

Notice that Jed travelled a total of 3 x 100 m = 300 m in the North direction, and 300 m + 100 m = 400 m in the East direction. Therefore the total distance he travelled is:  300 + 400 = 700 m.

But the actual displacement is given by the Pythagorean theorem as the hypotenuse of a right angle triangle of legs 300 m and 400 m:

displacement = \sqrt{300^2+400^2} =\sqrt{250000} =500\,m

5 0
3 years ago
In the reaction 2Ca + O2 → 2CaO for every 2 Ca you will need how much O2?
Serjik [45]

Answer:

1

Explanation: that is the ratio

5 0
2 years ago
An object with a mass of 20 kg has a net force of 80 N acting on it. What is the acceleration of the object?
Archy [21]

Answer:

4m/s^2

Explanation:

mass(m)=20 kg

force=80 N

acceleration (a)=?

Therefore,

Force = mass * acceleration

80 = 20*a

a=80/20

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7 0
2 years ago
Can someone help me?​
Leviafan [203]

Car X traveled 3d distance in t time.  Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t,  the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

V_y.t = \frac{2d}{t} .t = 2d

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

4 0
3 years ago
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