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3 years ago

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

Physics

2 answers:

Allisa [31]3 years ago

6 0

Answer:

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.

Read more on Brainly.com - brainly.com/question/788447#readmore

Explanation:

Amanda [17]3 years ago

4 0

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.

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A stone is thrown vertically into the air at an initial velocity of 96 ft/s. On Mars, the height s (in feet) of the stone above

vladimir1956 [14]

Answer:

240 ft

Explanation:

t = Time taken

u = Initial velocity = 96 ft/s

v = Final velocity

s = Displacement

a = Acceleration = 12 m/s² on Mars 32 ft/s² on Earth negative due to upward direction

Mars

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -12\times t^2\\\Rightarrow s=96t-6t^2\ ft$

Earth

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -32\times t^2\\\Rightarrow s=96t-16t^2\ ft$

Differentiating the first equation with respect to time we get

$\frac{ds}{dt}=96-12t$

Equating with zero

$0=96-12t\\\Rightarrow t=\frac{96}{12}=8\ s$

Differentiating the second equation with respect to time we get

$\frac{ds}{dt}=96-32t$

Equating with zero

$0=96-32t\\\Rightarrow t=\frac{96}{32}=3\ s$

Applying the time taken to the above equations, we get

$s=96t-6t^2\ ft\\\Rightarrow s=96\times 8-6\times 8^2\\\Rightarrow s=384$

$s=96t-16t^2\\\Rightarrow s=96\times 3-16\times 3^2\\\Rightarrow s=144$

Difference in height = 384-144 = 240 ft

The stone will travel 240 ft higher on Mars

6 0

3 years ago

Can someone help??

Paul [167]

Answer:

F = 2520 N

Explanation:

We have,

The maximum acceleration of a fist in a karate punch is 4200 m/s². The mass of the fist is 0.6 kg.

It is required to find the force that the wood place on the fist. Force is given by the product of mass and its acceleration such that,

F = ma

$F=0.6\times 4200\\\\F=2520\ N$

So, the force of 2520 N is acting on the wood.

8 0

3 years ago

a transmission-line cable, of length 3 km, consists of 19 strands of identical copper conductors, each 1.5 mm in diameter. becau

Ivan

the resistance of the cable is 582.9 ohms

we are given the length of the cable which is 3 km, of 1.5 mm in, the diameter and resistivity of copper which is 1.72 m

The formula we are referring to for calculating the resistance of the cable is

R = ρl/A.

As there are 19 strands of copper conductors, so the resistance will be

R = 19( ρl/A)

Here ρ is the resisitivity = 1.72 , l is the length = 3(1+0.05)*10³3= 3150 m

A=pie/4(1.5 x 10⁻³)^2 =1.766 x 10⁻⁶ =1.766 x 10^-6

Substituting the values in the formula we get

R = 19 ( 1.72*3150 )/1.766 x 10⁻⁶

= 582.9 ohm

To know more about resistance refer to the linkhttps://brainly.com/question/14547003?referrer=searchResults.

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1 year ago

I feel really isolated these days and wish I had more friends. I've chosen to put myself in new social situations to try to make

leva [86]

Behaviorist approach to psychology

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3 years ago

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$

$0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

PART B) Replacing the values given as,

$h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

$v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}$

$v_f = 1.8486m/s$

Therefore the speed of the masses would be 1.8486m/s

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3 years ago