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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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Joel uses a claw hammer to remove a nail from a wall. He applies a force of 40 newtons on the hammer. The hammer applies a force
jarptica [38.1K]

Hi!


Mechanical advantage is defined as the<em> ratio of force produced by an object to the force that is applied to it.</em>

In our case, this would be the ratio of the force applied by the claw hammer on the nail to the force Joel applies to the claw hammer, which is

160:40 or 4:1

So the mechanical advantage of the hammer is four.


Hope this helps!


3 0
2 years ago
Read 2 more answers
The boom hoisting sheave must have pitch diameters of no less than _______times the nominal diameter of the rope used.
alexira [117]

Answer:

18 times

Explanation:

According to the security purposes which is set under the rules and regulation OSHA, which describes all the rights to the worker.

In the boom hoist receiving system all the sheaves which are used should have a pitch diameter of rope not less than 18 times the diameter of the nominal rope which is used.

7 0
2 years ago
What is the net force required to give an automobile with a mass of 1,600 kg an acceleration of 4.5 m/s2
Paladinen [302]

Answer:

Fnet=7200 N

Explanation:

Fnet=mass x acceleration

mass= 1600kg acceleration=4.5m/s^2

8 0
2 years ago
A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direc
Alborosie

The given question is incomplete. The complete question is as follows.

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

A) If the net force acting on the particle is 6.21 \times 10^{-3} N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.

Enter your answers numerically separated by commas.

Explanation:

The given data is as follows.

           Q = 6.50 \times 10^{-6} C

           E = 1300 N/C in the +x direction

           B = 1.02 T in the +z direction

and,    F_{net} = 6.25 \times 10^{-3} N in the +x direction

Also,       F_{net} = F_{E} - F_{b}

                         = qE - qvB

Now, we will calculate the value of v as follows.

             v = (\frac{1}{B}) \times (E - \frac{F_{net}}{q})

                 = (\frac{1}{1.02 T}) \times (1300 - \frac{6.25 \times 10^{-3}}{6.50 \times 10^{-6}})

                v = 458.507 m/s

Using the value for velocity, we need to know which direction it's going.

You know +x direction for E, +z direction for B and +x for F_{net}.

Using the right hand rule where:

your right thumb goes toward the F_{net}, then your index finger points to B (z direction) Then curl your middle, ring, and pink 90 angle. This shows where v is going which is -y direction.

Thus, we can conclude that v_{x}, v_{y}, v_{z} = 0, -(458.507), 0.

8 0
2 years ago
calculate the workdone to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it
Lera25 [3.4K]
The force of F=10 N produces an extension of
x=4 cm=0.04 m
on the string, so the spring constant is equal to
k= \frac{F}{x}= \frac{10 N}{0.04 m}=250 N/m

Then the string is stretched by \Delta x=40 cm=0.40 m. The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
W= \Delta U= \frac{1}{2}k(\Delta x)^2  = \frac{1}{2}(250 N/m)(0.40 m)^2=20 J
5 0
2 years ago
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