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SashulF [63]
4 years ago
13

Which of the following are examples of levers?

Physics
1 answer:
guajiro [1.7K]4 years ago
5 0

The examples of levers are seesaw, door and corkscrew

Answer: Option A, B, and C

<u>Explanation:</u>

A simple machine which is capable of rotating or moving with respect to a fixed point at the center of mass is termed as lever. Lever is made up of a rigid rod which acts as the beam or the object on which the effort can be applied based on the load and a fixed hinge like thing named as fulcrum or the center of mass.

The fulcrum will be fixed and the beam will be rotating or moving with the support of fulcrum at the center for given load or effort. So among the given options, ramp is not falling under the categories of lever as door also have a fulcrum. But the ramp is only an inclined plane without fulcrum. But the Seesaw, door and the corkscrew has both fulcrum and beam which confirms them falling in the category of lever.

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Which of the following is not an example of kinetic energy being converted to potential energy?
KengaRu [80]

The list of choices you provided with your question
is utterly devoid of any such examples.

6 0
4 years ago
Read 2 more answers
13. Austin rode his bike 10 m/s for two minutes. How far did he travel? A. 200 meters B. 1200 meters C. 1000 meters D. 20 meters
Semmy [17]

Answer:

B. 1200

Explanation:

60 sec in one min in 2 min there will be 120 sec. 10x120=1200

5 0
3 years ago
Calculate the gravitational attraction between two objects which masses are: mass A: 2.5kg, mass B: 5kg. The distance between th
jonny [76]

From the calculation, the gravitational force of attraction is  1.33 * 10^-14 N.

<h3>What is the gravitational force?</h3>

The gravitational force is an attractive force that acts between any two masses.

It is given by;

F = Gm1m2/r^2

F = 6.67 * × 10−11 * 2.5 * 5/(250)^2

F  = 83.4  × 10−11 /62500

F= 1.33 * 10^-14 N

Learn more about gravitational force:brainly.com/question/12528243

#SPJ1

4 0
2 years ago
What is the highest degrees above the horizon the moon ever gets during the year in the Yakima Valley ?
Ivahew [28]

The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.

Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima.  The latitude in the middle
of that intersection is 46.585° North.  <u>That's</u> the number we need.

Here's how I would do it:

-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.

-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°. 

-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.

This sets the limit of the highest in the sky that the moon can ever appear.

90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .

That doesn't happen regularly.  It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).

Depending on the time of year, that can be any time of the day or night.

The most striking combination is at midnight, within a day or two of the
Winter solstice, when the moon happens to be full.

In general, the Full Moon closest to the Winter solstice is going to be
the moon highest in the sky.  Then it's going to be somewhere near
67° above the horizon at midnight.


5 0
3 years ago
A coin is thrown with a velocity of 0 m/s down a dry well and hits bottom in 1.2s, what’s the depth of the well?
pentagon [3]

Answer:

The well is 7.1 meters deep.

Explanation:

The formula to use here is the distance in a uniformly accelerated motion:

d = \frac{1}{2}at^2+v_0t+d_0

where d stands for distance, t for time, a for acceleration, v0 and d0 for initial velocity and distance, respectively. Since the initial distance and velocity are both zero, we are left with the first term. The coin is in free fall and so it is accelerated by gravity:

d = \frac{1}{2}at^2= \frac{1}{2}gt^2=\frac{1}{2}9.8\frac{m}{s^2}1.2^2s^2=7.1m

The well is 7.1 meters deep.

5 0
4 years ago
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