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3 years ago

Hi! Please help me!!!!! I don’t understand this question at all. If you could please explain that would be so helpful! Thanks!

Physics

2 answers:

inessss [21]3 years ago

7 0

Answer:

I would say the stone is being used as a fulcrum.

Explanation:

An example of a fulcrum is the triangle thing under a seesaw. I don't know for sure if that's the right answer, but I would strongly suggest it. Hope I helped, have a nice day :)

Cerrena [4.2K]3 years ago

6 0

Answer:

to lift and balance an object, the effort force the user applies multiplied by its distance to the fulcrum must equal the load force multiplied by its distance to the fulcrum. Consequently, the greater the distance between the effort force and the fulcrum, the heavier a load can be lifted with the same effort force.

Explanation:

Have you ever wondered how ancient people could lift very heavy objects, such as large stones, to build pyramids? A lever is a simple machine that can help people do just this. It can also help make other kinds of physical work easier by giving the user a mechanical advantage.

Common examples of levers you might see around you are seesaws, scissors, wheelbarrows and even the your own jaw. Although all of these levers have the same functional parts, they vary in where the different components are located. How much effort does it take to lift a heavy load using a common type of lever.

Seesaws and scissors belong to a certain class of levers, called class 1. Class 1 levers usually have a beam that is rigid, long and thin, like a ruler. Between the two ends of the beam is the fulcrum, or pivot point, which is the point at which the beam can balance and move freely up and down around. On one end, the user places the load to be moved. On the other end, the user can apply effort, or a force, to try and move the load.

The way levers work is by multiplying the effort exerted by the user. Specifically, to lift and balance an object, the effort force the user applies multiplied by its distance to the fulcrum must equal the load force multiplied by its distance to the fulcrum. Consequently, the greater the distance between the effort force and the fulcrum, the heavier a load can be lifted with the same effort force.

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You are designing a hydraulic lift for a machine shop. The average mass of a car it needs to lift is about 1500 kg. You wish to

ehidna [41]

Answer:

(a) Area(small piston)/Area(large piston) = 0.037

(b) h = 1336.36 cm = 13.36 m

Explanation:

(a)

The stress on the smaller piston is equally transmitted to the larger piston, in a hydraulic lift. Therefore,

Stress (small piston) = Stress (large piston)

Force (small piston)/Area (small piston) = Force (Large Piston)/Area (Large Piston)

Area(small piston)/Area(large piston) = Force (small piston)/Force(Large piston)

Area(small piston)/Area(large piston) = 550 N/(1500 kg)(9.8 m/s²)

Area(small piston)/Area(large piston) = 0.037

(b)

The work is also transmitted equally to the large piston. So,

Work(small piston) = Work(Large Piston)

Force(small piston).Displacement(small piston) = Force(large piston).Displacement(small piston)

(550 N)(h) = (1500 kg)(9.8 m/s²)(50 cm)

h = 735000 N.cm/550 N

h = 1336.36 cm = 13.36 m

5 0

3 years ago

The 2 question are on the photo above.

AlekseyPX

A. power plants burn coal. A fossil-fuel power plant is one that burns fossil fuels such as coal, natural gas or petroleum (oil) to produce electricity.

b. Fossil fuels are called so because they have been derived from fossils, which were formed millions of years ago during the time of the dinosaurs. They are fossilized organic remains that over millions of years have been converted to oil, gas, and coal.

C. they are generally classified as non-renewable resources because they take millions of years to form and known viable reserves are being depleted much faster than new ones are generated.

5.
a. Gravitational potential energy and work done
If an object is lifted, work is done against the force of gravity.
When work is done energy is transferred to the object and it gains gravitational potential energy.
If the object falls from that height, the same amount of work would have to be done by the force of gravity to bring it back to the Earth’s surface.
If an object at a certain height has 2000 J of gravitational potential energy, we can say that:
2000 J of work has been done in getting the object to that height from the ground
and
2000 J of work would have to be done to bring it back to the ground.

3 0

3 years ago

The transmission coefficient of the hypothetical disease dumas fever is 0.2. if the combined death and recovery rate is 0.4, wha

borishaifa [10]

NT = r / β
where NT is the threshold density, r is the recovery and death rate and β is the transmission coefficient.

= 0.4 / 0.2
= 2

7 0

3 years ago

How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 n/m so that it wil

mrs_skeptik [129]

The frequency of a simple harmonic oscillator such as a spring-mass system is given by
$f= \frac{1}{2 \pi} \sqrt{ \frac{k}{m} }$
where
k is the spring constant
m is the mass attached to the spring.

Re-arranging the formula, we get:
$m= \frac{k}{4 \pi^2 f^2}$
and since we know the constant of the spring:
$k=39.5 N/m$
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
$m= \frac{39.5 Hz}{4 \pi^2 (1.00 Hz)^2} = 1.00 kg$

7 0

3 years ago

A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

$\theta = \omega t\\\\\theta = (100 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m$

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

6 0

3 years ago