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3 years ago

Second and third class levers are differentiated by __________.

Physics

2 answers:

ira [324]3 years ago

5 0

Answer:

i think it is b

hope this helps

Explanation:

azamat3 years ago

3 0

Answer: B

Explanation:the load in the second class lever is found in between the effort and the fulcrum

While in third class lever effort is found in between load and fulcrum

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To meet the minimum required instrument flight experience to act as a pilot in command of an aircraft under IFR, you must have l

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I think its a tbh bc it seems to be the best answer out of a b c and d

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3 years ago

The legal tradition that kept women from owning property and holding public office came to the United States from

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<span>The legal tradition that kept women from owning property and holding public office came to the United States from: C. Britain.</span>

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3 years ago

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Keisha is making a diagram of a simple machine in the body.

Amiraneli [1.4K]

Answer:

Sorry for being late. It is...

A.) X: Load, Y: Fulcrum, Z: Lever

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3 years ago

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Suppose that instead of being inclined to Earth's orbit around the Sun, the Moonâs orbit was in the same plane as Earthâs orbit

ycow [4]

Answer:

c) 12

Explanation:

A Solar eclipse occurs when The Sun, The Earth and The Moon comes in a straight line with the Moon being in between the Earth and the Sun. At this point the Moon appears to block the Sun and Moon's shadow falls on Earth. This would occur only on the day of the New Moon.

If the Moon's orbit was in the same plane as that of the Earth's orbit. Every new Moon, there would be a Solar Eclipse. The Lunar cycle is of 29.5 Days which means there will be one new Moon every month. So there will be 12 Solar Eclipses every year.

Currently, the orbit of the Moon is tilted at an angle of 5° thus we don't see that many Solar eclipses. Maximum of 5 solar eclipses can occur in an year.

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2 years ago

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode

sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell $v_{0}$ in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

$v_{x}=v_{0} \times \cos \theta$

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of $v_{x}$ and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

$v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}$

$v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}$

For motion with constant acceleration, we know

$s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}$

Along the horizontal, x-axis, we might write this as

$x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}$

Measuring distances relative to the firing point means

$x_{0}=0$

we know that,

$a_{x}=0$

or,

$v_{x}=v_{x 0}=\text { constant }$

By applying the values, we get,

$x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}$

The acceleration of gravity is vertically downward and is $g=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

$y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}$

we know, $y_{0}=0$ and $a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , so,

$y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)$

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

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3 years ago