(03.02 LC)

The Northern Hemisphere is warmer in spring than in winter, because in spring

Amanda [17]

Both hemispheres are warmer in their Spring than they are
in their Winter, because . . .

A). the sun climbs higher in the sky during Spring than it does during
WInter ... shooting its rays more directly at the ground ...,

and

B). the sun stays up in the sky longer in Spring than it does in
Winter, giving the ground more time to absorb its rays.

4 0

4 years ago

Read 2 more answers

Compare and contrast carbon to its isotopes: what is different about them and what is the same?

Alex_Xolod [135]

Their atomic numbers are all the same, because they all have the same number of protons. But their atomic weights are different, because they have different numbers of neutrons.

6 0

4 years ago

A 7.25 kg7.25 kg block is sent up a ramp inclined at an angle ????=28.5°θ=28.5° from the horizontal. It is given an initial velo

Free_Kalibri [48]

Answer:

15.03 m

Explanation:

Given:

mass of the block, m = 7.25 kg

Angle, Θ = 28.5°

Initial speed of the block, v₀ = 15 m/s

let the distance traveled by the block be 's'

Now, applying the work energy theorem,

we have

$(m\times g\times\sin(\theta)\times s) + \mu_k\times mg\times s\times cos(\theta) = \frac{1}{2}\times m\times v^2$

on substituting the values in the above equation, we get

$(7.25\times 9.8\times\sin(28.5^o)\times s) + 0.326\times 7.25\times9.8\times s\times cos(28.5^o) = \frac{1}{2}\times 7.25\times 15^2$

or

$33.902\times s) +20.35\times s = 815.625$

or

$54.252\times s = 815.625$

s = 15.03 m

Hence, the block will travel 15.03 m up the ramp

6 0

3 years ago

Suppose a 1 Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to Ear

vagabundo [1.1K]

Answer:

The time required to send the data from Earth to Moon will be 1.28s while for a two way communication, to send it back to the earth, it will take double time i.e. RTT = 2.56s

Explanation:

Distance between Earth and Moon = 385,000 km = 3.85 x 10⁸m

Speed of data travel = speed of light ≈ 3 x 10⁸m/s

As, v=d/t

t=d/v

$t=\frac{3.85*10^{8} }{3*10^{8}}$

t=1.28s

RTT = Double of single way time taken = 2x1.28

RTT=2.56s

7 0

3 years ago

Rank the deformations of the following rods in terms of the magnitude of the average normal strain: (a) The length of a 1-m-long

kipiarov [429]

To solve this problem we will consider the concepts related to the normal deformation on a surface, generated when the change in length is taken per unit of established length, that is, the division between the longitudinal fraction gained or lost, over the initial length. In general mode this normal deformation can be defined as

$\epsilon = \frac{\delta}{l} = \frac{l_0-l}{l}$