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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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A moving roller coaster speeds up with constant acceleration for 2.3s until it reaches a velocity of 35m/s. During this time, th
nata0808 [166]
Hope this answer helps: option B

Explanation:

9 0
3 years ago
Velocity question...is it A B Cor D
Ket [755]
In a constant acceleration of 3m per second, after 10 seconds,

3 x 10 = 30

B. 30m/s is your answer

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6 0
2 years ago
Read 2 more answers
A 5.0-kg centrifuge takes 95 s to spin up from rest to its final angular speed with constant angular acceleration. A point locat
stellarik [79]

Answer:

(a) 17.37 rad/s^2

(b) 12479

Explanation:

t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0

w = v / r = 99 / 0.06 = 1650 rad/s

(a) Use first equation of motion for rotational motion

w = w0 + α t

1650 = 0 + α x 95

α = 17.37 rad/s^2

(b) Let θ be the angular displacement

Use third equation of motion for rotational motion

w^2 = w0^2 + 2 α θ

1650^2 = 0 + 2 x 17.37 x θ

θ = 78367.87 rad

number of revolutions, n = θ / 2 π

n = 78367.87 / ( 2 x 3.14)

n = 12478.9 ≈ 12479

4 0
3 years ago
Could anyone help me for this question please!! Really emergency!!!
kodGreya [7K]
16/9 m/s^2
negative 4/3 m/s^2
14 m/s
the last one is too detailed to do in my head while on the bus; sorry
5 0
3 years ago
A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?
anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\  \end{gathered}

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}

now, replace

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}

hence, the answer is 0.495 m/s

3 0
11 months ago
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