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sweet-ann [11.9K]
2 years ago
5

When Earth’s Northern Hemisphere is tilted toward the Sun during June, some would argue that the cause of our seasons is that th

e Northern Hemisphere is physically closer to the Sun than the Southern Hemisphere, and this is the primary reason the Northern Hemisphere is warmer. What argument or line of evidence could contradict this idea?
Physics
1 answer:
Ludmilka [50]2 years ago
4 0

Answer:

Distance of Earth from the Sun has nothing to do with the seasons only the tilt is responsible for the change in seasons.

Explanation:

The Earth's tilt does cause the seasons but the distance from the sun and has nothing to do with the change in seasons. In June, when the Northern Hemisphere is tilted in the direction of the Sun during the Northern Hemisphere summer the Earth is actually farthest from the Sun. In January, when the Southern Hemisphere is tilted in the direction of the Sun during the Northern Hemisphere winter the Earth is actually closest to the Sun. This is caused due to the elliptical orbit of the Earth. So, distance of Earth from the Sun has nothing to do with the seasons.

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How much force is required to pull a spring 3.0 cm from
avanturin [10]

Answer:

I know that T= kx where T is the tension which equaka the force og gravity = mg = 1.37 * 10 = 13.7 x is the elongation of the spring so the length after dangling the object minus the original length.

I hope it helps

plz let me know if it is wrong or right.

4 0
2 years ago
A balloon drifts 140m toward the west in 45s ; then the wind suddenly changes and the balloon flies 90m toward the east in the n
Bogdan [553]

Answer: 140 m

Explanation:

Let's begin by stating clear that motiont is the change of position of a body at a certain time. So, during this motion, the balloon will have a trajectory and a displacement, being both different:

The<u> trajectory</u> is <u>the path followed by the body, the distance it travelled</u> (is a scalar quantity).  

The displacement is <u>the distance in a straight line between the initial and final position</u> (is a vector quantity).  

So, according to this, the distance the balloon traveled during the first 45 s (its trajectory) is 140 m.

But, if we talk about displacement, we have to draw a straight line between the initial position of the balloon (point 0) to its final position (point 90 m).  Being its displacement 95 m.

8 0
1 year ago
An object thrown in the air has a velocity after t seconds that can be described by v(t) = -9.8t + 24 (in meters/second) and a h
marin [14]

Answer and Explanation: Kinetic energy is related to movement: it is the energy an object possesses during the movement. it is calculated as:

K=\frac{1}{2}mv^{2}

For the object thrown in the air:

K=\frac{1}{2}.2.[v(t)]^{2}

K=(-9.8t+24)^{2}

K=96.04t^{2}-470.4t+576

Kinetic energy of the object as a function of time: K=96.04t^{2}-470.4t+576

Potential energy is the energy an object possesses due to its position in relation to other objects. It is calculated as:

U=mgh

For the object thrown in the air:

U=9.8.2.h(t)

U=9.8.2.(-4.9t^{2}+24t+60)

U=-96.04t^{2}+470.4t+1176

Potential energy as function of time: U=-96.04t^{2}+470.4t+1176

Total kinetic and potential energy, also known as mechanical energy is

TME = 96.04t^{2}-470.4t+576 + (-96.04t^{2}+470.4t+1176)

TME = 1752

The expression shows that total energy of an object thrown in the air is constant and independent of time.

6 0
1 year ago
A certain signal molecule S in heart tissue is degraded by two different biochemical pathways: when only Path 1 is active, the h
Misha Larkins [42]

Answer:

Half life of S = 3.76secs

Explanation:

The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.

For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564

for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615

Simply say = 1/3.9 + 1/104 = 0.266015

So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active

6 0
2 years ago
8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres
Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

hence, B_o = 1.013μT

6 0
2 years ago
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