• A flame always points upwards. Why do you think this is so?

En una barra de 6m que se utiliza como palanca se coloca el fulcro a 2 m de distancia del extremo derecho, como se muestra en la

Anna007 [38]

Answer:

Fp = 45 N

VMR = 2

VMI = 2

Efficiency = 100%

Explanation:

English Translation

In a 6m bar that is used as a lever, the fulcrum is placed 2m away from the right end, as shown in the figure. At that same end it is required to support a load of 90N. Neglecting lever weight, determined Fp, VMI, VMR and system efficiency

Solution

With the diagram not massively needed to solve this question.

Although, the load of 90 N is placed at the right end of the bar, the fulcrum is placed 2 m from that right end and the balancing force or effort is placed at the extreme most left end of the 6 m bar.

Taking moment about the fulcrum,

The sum of clockwise moments must balance the sum of anti-clockwise moments.

(Load) × (Distance of load from the fulcrum) = (Balancing force) × (Distance of balancing force from fulcrum)

Load = L = 90 N

Distance of load from the fulcrum = 2 m

Balancing force or Effort = Fp = ?

Distance of balancing force from fulcrum = 6 - 2 = 4 m

90 × 2 = Fp × 4

Fp = (180/4) = 45 N

The mechanical advantages are them given as

VMR = (Load)/(Effort) = (90/45) = 2

VMI = (Distance of Effort from the fulcrum)/(Distance of Load from fulcrum) = (4/2) = 2

Efficiency is then given as the VMR divided by VMI in percentage terms

Efficiency = 100% × (VMR/VMI)

Efficiency = 100% × (2/2) = 100%

Hope this Helps!!!

4 0

3 years ago

What did the scientist Robert Hooke claim ?

trapecia [35]

Robert Hooke (1635 - 1703) The Englishman Robert Hooke (18th July 1635 - 3rd March 1703) was an architect, natural philosopher and brilliant scientist, best known for his law of elasticity (Hooke's law), his book Micrographia, published in 1665 and for first applying the word "cell" to describe the basic unit of life.

7 0

3 years ago

Read 2 more answers

A proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.

lawyer [7]

The change in potential energy of the proton is 5.6 x $10^{-17}$ Joule

<h3>What is a Uniform Electric Field ?</h3>

A uniform electric field is where the electric field strength is the same at all points in the field. In the uniform field, the force experienced by a charge is the same no matter where it is placed in the field.

Given that a proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.

The distance d = 10 cm = 0.1 m

Electric field E = 3.5 KN/C

Proton charge q = 1.6 x $10^{-19}$ C

The Work done = Fd

but F = Eq

Recall that Electric field E = F/q = V/d

Where V = potential difference.

Let us first calculate the V

E = V/d

V = Ed

Substitute all the parameters into the formula above

V = 3.5 × 10³ × 0.1

V = 350 v

from F/q = V/d

make F the subject of formula and substitute it in work formula

F = Vq/d

W.D = Vq/d x d

W.D = Vq

Substitute all the parameters into the formula above

W.D = 350 x 1.6 x $10^{-19}$

W.D = 5.6 x $10^{-17}$ J

Work done = Energy = Potential Energy

Therefore, the change in potential energy of the proton is 5.6 x $10^{-17}$ Joule

Learn more about Electric Field here: brainly.com/question/14372859

#SPJ1

7 0

2 years ago

A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f

amid [387]

You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude W = m g

• the normal force, pointing perpendicular to the hill and away from the ground with mag. N

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector F pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. F.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = W (//) = m a (//)

∑ (⟂) = W (⟂) + N = m a (⟂)

where, for instance, W (//) denotes the component of the sled's weight in the direction parallel to the hill, while a (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -F to the first equation.

If the hill makes an angle of θ with flat ground, then W makes the same angle with the hill so that

W (//) = -m g sin(θ)

W (⟂) = -m g cos(θ)

So we have

-m g sin(θ) = m a (//) → a (//) = -g sin(θ)

-m g cos(θ) + N = m a (⟂) → a (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

a = a (//)

with magnitude

||a|| = a = g sin(θ).

6 0

3 years ago

What would the weight of an astronaut be on Saturn if his mass is 68 kg and acceleration of gravity on Saturn is 10.44 m/s2? Ple

alex41 [277]

Here's the part you need to know:

   (Weight of anything) =

   (the thing's mass)
times
       (acceleration of gravity in the place where the thing is) .

   Weight = (mass ) x (gravity) .

That's always true everywhere.
You should memorize it.

For the astronaut on Saturn . . .

   Weight = (mass ) x (gravity) .

   Weight = (68 kg) x (10.44 m/s²)

   = 709.92 newtons .
__________________________________

On Earth, gravity is only 9.8 m/s².
So as long as the astronaut is on Earth, his weight is only

   (68 kg) x (9.8 m/s²)

   = 666.4 newtons .

Notice that his mass is his mass ... it doesn't change
no matter where he goes.

But his weight changes in different places, because
it depends on the gravity in each place.

4 0

3 years ago