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Butoxors [25]
1 year ago
10

You use a lever to lift a heavy tree branch you apply a force of 50 n and the lever lifts the branch

Physics
1 answer:
valentinak56 [21]1 year ago
5 0

1.8 is the mechanical advantage of the lever.

<h3>Definition of mechanical advantage</h3>

The theoretical mechanical advantage of a system is the ratio of the force that performs the useful work to the force applied, assuming there is no friction in the system.

The advantage gained by the use of a mechanism in transmitting force specifically the ratio of the force that performs the useful work of a machine to the force that is applied to the machine.

Mechanical advantage is given by the ratio of the load lifted to the force applied to lift the load.

In this case, Mechanical advantage=L/E where L is the load and E is the effort applied.

Mechanical advantage= 90/50 =1.8

Question-you use a lever to lift a heavy tree branch. you apply a force of 50 n and the lever lifts the branch with a force of 90 n. what is the mechanical advantage of the lever?

To learn more about the Mechanical advantage visit the link

brainly.com/question/16617083

#SPJ4

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Three laws that indicate a chemical change
alexandr1967 [171]

Temperature, The highness, and the time.

Hope this helps!

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4 0
3 years ago
How much heat is needed to raise the temperature of an empty 2.0 x 101 kg vat made of
scZoUnD [109]

Answer:

Q = 7272 Kilojoules.

Explanation:

<u>Given the following data;</u>

Mass = 2.0*101kg = 202kg

Initial temperature, T1 = 10°C

Final temperature, T2 = 90°C

We know that the specific heat capacity of iron = 450J/kg°C

*To find the quantity of heat*

Heat capacity is given by the formula;

Q = mcdt

Where;

  • Q represents the heat capacity or quantity of heat.
  • m represents the mass of an object.
  • c represents the specific heat capacity of water.
  • dt represents the change in temperature.

dt = T2 - T1

dt = 90 - 10

dt = 80°C

Substituting the values into the equation, we have;

Q = 202*450*80

Q = 7272KJ or 7272000 Joules.

6 0
3 years ago
the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path
Svetach [21]

Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:  

V=\sqrt{G\frac{M}{r}}

Where:  

V is the speed of travel of the Moon around the Earth

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24} kg is the mass of the Earth

r=384400(10)^{3} m is the distance from the center of the Earth to the center of the Moon

Solving:

V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{384400(10)^{3} m}}

V=1018.26 m/s This is the speed of travel of the Moon around the Earth

5 0
3 years ago
when the mass of an object is increased it would decrease its acceleration in the force is left alone
mojhsa [17]

Answer:

See the explanation below.

Explanation:

This analysis can be easily deduced by means of Newton's second law which tells us that the sum of the forces or the total force on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = total force [N]

m = mass [kg]

a = acceleration [m/s²]

We must clear the acceleration value.

a=\frac{F}{m}

We see that the term of the mass is in the denominator, so that if the value of the mass is increased the acceleration decreases, since they are inversely proportional.

4 0
3 years ago
Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible co
Illusion [34]

Answer:

 q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

          F_e -F_g = 0

          F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

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In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

          k q² = G m²

          q = \sqrt{ \frac{G}{k} }    m

we substitute

           q = \sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }   m

            q = \sqrt{0.7419 \ 10^{-20}}   m

           q = 0.861 10⁻¹⁰ m

           q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

            q = 8.61 10⁻¹¹ 10³

            q = 8.61 10⁻⁸ C

             

this is a very small charge value so it should be easy to create in each one

6 0
3 years ago
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