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garik1379 [7]
3 years ago
7

Second and third class levers both have __________.

Physics
2 answers:
Zolol [24]3 years ago
5 0

Answer:

i think its d on edgenuity

Explanation:

nataly862011 [7]3 years ago
4 0

The answer would be:

<u><em>D. the fulcrum at one end of the lever.</em></u>

Both second class and third class levers have their fulcrum located at the end of the lever. They differ in the location of the load and effort. For third class levers the effort is in between the load and the fulcrum. For second class levers, the load is in between the effort and fulcrum. You can look at the attached picture as reference.

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At what speed does a 1,248 kg compact car have the same kinetic energy as a 18,777 kg truck going 26 km/h?
notka56 [123]

The speed of car is 100.8km/h

KE =  \frac{1}{2} m(v \: truck ) {}^{2}

= 0.5 \times 18777 \times  \frac{26}{3.6} \times  \frac{26}{3.6}

= 489708.79j

=  \frac{1}{2}  \times 1248 \times( v \: car) {}^{2}

= 624v {}^{2}

so \: v {}^{2}  =  \frac{489708.7}{624}

= 784

v =  \sqrt{784}

v = 28m/s

v car= 28×3.6

=100.8km/h

Hence, the speed of the car is 100.8km/h

learn more about speed from here:

brainly.com/question/28326855

#SPJ4

3 0
1 year ago
What number on this diagram refers to the tick marks for the dependent variable?
givi [52]

Answer:

I'm pretty sure it's 3.

Explanation:

Because if you look at your options the only that would be relevant to tick marks would be either 4 or 3. And it said in the question that we're looking for the one for the dependent variable. And the dependent variable is on the Y- Axis and the 3 is the tick marks for the y-axis. So your answer is 3.

4 0
3 years ago
after a large snowstorm you shovel 2000 kilograms of snow off your side walk in 1 hour. you lift the shovel to an average height
solmaris [256]

Lifting a mass to a height, you give it gravitational potential energy of

       (mass) x (gravity) x (height)  joules.

To give it that much energy, that's how much work you do on it.

If 2,000 kg gets lifted to 1.25 meters off the ground, its potential energy is

         (2,000) x (9.8) x (1.25) = 24,500 joules.

If you do it in 1 hour (3,600 seconds), then the average power is

           (24,500 joules) / (3,600 seconds) = 6.8 watts.

None of these figures depends on whether the load gets lifted all at once,
or one shovel at a time, or one flake at a time.

But this certainly is NOT all the work you do.  When you get a shovelful
of snow 1.25 meters off the ground, you don't drop it and walk away, and
it doesn't just float there. You typically toss it, away from where it was laying
and over onto a pile in a place where you don't care if there's a pile of snow
there. In order to toss it, you give it some kinetic energy, so that it'll continue
to sail over to the pile when it leaves the shovel.  All of that kinetic energy
must also come from work that you do ... nobody else is going to take it
from you and toss it onto the pile.


8 0
3 years ago
A ray of light is incident on a body x what is the reftactive index​
evablogger [386]

Answer:

1.63

Explanation:

If you have the following options:

<u>A.  1.63</u>

B.  1.50

C.  1.49

D.  1.33

E.  1.02

6 0
3 years ago
A block is attached to a spring, with spring constant k, which is attached to a wall. it is initially moved to the left a distan
Ghella [55]

Answer:

x(t) = d*cos ( wt )

w = √(k/m)

Explanation:

Given:-

- The mass of block = m

- The spring constant = k

- The initial displacement = xi = d

Find:-

- The expression for displacement (x) as function of time (t).

Solution:-

- Consider the block as system which is initially displaced with amount (x = d) to left and then released from rest over a frictionless surface and undergoes SHM. There is only one force acting on the block i.e restoring force of the spring F = -kx in opposite direction to the motion.

- We apply the Newton's equation of motion in horizontal direction.

                             F = ma

                             -kx = ma

                             -kx = mx''

                              mx'' + kx = 0

- Solve the Auxiliary equation for the ODE above:

                              ms^2 + k = 0

                              s^2 + (k/m) = 0

                              s = +/- √(k/m) i = +/- w i

- The complementary solution for complex roots is:

                              x(t) = [ A*cos ( wt ) + B*sin ( wt ) ]

- The given initial conditions are:

                              x(0) = d

                              d = [ A*cos ( 0 ) + B*sin ( 0 ) ]

                              d = A

                              x'(0) = 0

                              x'(t) = -Aw*sin (wt) + Bw*cos(wt)

                              0 = -Aw*sin (0) + Bw*cos(0)

                              B = 0

- The required displacement-time relationship for SHM:

                               x(t) = d*cos ( wt )

                               w = √(k/m)

3 0
3 years ago
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