Ask question

Ask question

All categories

3 years ago

5

On a particular day, Denali had a temperature of -23°C at its peak and 13°C at its base. What could be the reason for this tempe

rature difference?
A. temperature increases with an increase in altitude
B.temperature decreases with an increase in altitude
C. the base is close to the gulf of Alaska
D.the mountain receives lake effect snow

1 answer:

ozzi3 years ago

3 0

I think the answer is A. Hope this helped?

You might be interested in

Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of

Alchen [17]

Answer:

$\theta=10.20^{\circ}$

$\Delta l=0.10 ft$

Explanation:

First of all, we analyze the system of blocks before starting to move.

$\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0$

$\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0$

$11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0$

$11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0$

$11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0$

$16sin(\theta)-2.91cos(\theta)=0$

$tan(\theta)=0.18$

$\theta=arctan(0.18)$

$\theta=10.20^{\circ}$

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.

$P_{A}sin(\theta)-F_{fA}-F_{spring}=0$

Where:

$F_{spring} = k\Delta l=2.1\Delta l$

$P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0$

$\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}$

$\Delta l=0.10 ft$

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0

3 years ago

Suppose an empty grocery cart rolls downhill in a parking lot. The cart has a maximum speed of 1.3 m/s when it hits the side of

Ket [755]

The cart comes to rest from 1.3 m/s in a matter of 0.30 s, so it undergoes an acceleration <em>a</em> of

<em>a</em> = (0 - 1.3 m/s) / (0.30 s)

<em>a</em> ≈ -4.33 m/s²

This acceleration is applied by a force of -65 N, i.e. a force of 65 N that opposes the cart's motion downhill. So the cart has a mass <em>m</em> such that

-65 N = <em>m</em> (-4.33 m/s²)

<em>m</em> = 15 kg

4 0

3 years ago

Bone has a Young's modulus of about

saveliy_v [14]

Answer: 4.74 mm

Explanation:

We can solve this problem with the following equation:

$Y=\frac{stress}{strain}$ (1)

Where:

$Y=1.8(10)^{10} Pa$ is the Young modulus for femur

$stress=\frac{F}{A}=1.58(10)^{8} Pa$ is the stress (force $F$ applied per unit of transversal area $A$ ) on the femur

$strain=\frac{\Delta l}{l_{o}}$

Being:

$\Delta l$ the compression the femur can withstand before breaking

$l_{o}=0.54 m$ is the length of the femur without compression

Writing the data in equation (1):

$Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l_{o}}}$ (2)

$1.8(10)^{10} Pa=\frac{1.58(10)^{8} Pa}{\frac{\Delta l}{0.54 m}}$ (3)

Isolating $\Delta l$ :

$\Delta l=\frac{(1.58(10)^{8} Pa)(0.54 m)}{1.8(10)^{10} Pa}$ (4)

$\Delta l=0.00474 m$ (5) This is the compression in meters

Converting this result to millimeters:

$\Delta l=0.00474 m \frac{1000 mm}{1 m}=4.74 mm$

4 0

3 years ago

If an object is rolling without slipping, how does its linear speed compare to its rotational speed?

RoseWind [281]

Answer:

v = rw

Explanation:

When an object is rolling continuously without slipping, then every angle it rotates through, is equal to a distance the perimeter has rotated.

If the object completes 10 revolutions and takes a particular time, let's say t to complete it. The angular distance would then be 20 π rad, while its angular velocity will be 20 π/t

The circumference will somehow translate to the distance it covers, which is 20πr, this means that the speed is 20πr/t

So, like the question asked, the linear speed compared to angular speed is

v : w

20πr/t : 20πt, which can be simplified to

r : 1

In essence, v = rw

3 0

4 years ago

If you plan to exercise every day what is something you can do to make it more convenient so that you exercise more regularly?

jolli1 [7]

Answer: exercise doing something you like. for example if you like dancing incorporate dancing in your workouts so you will be more likely to do it regularly

Explanation:

4 0

3 years ago