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3 years ago

. How much work is done in lifting an object that has a mass of 5kg a vertical distance of 2m?

Physics

1 answer:

malfutka [58]3 years ago

7 0

<h3>Answer:</h3>

100 Joules

<h3>Explanation:</h3>

The question is testing on how to calculate work done;

We are given;

Mass of an object is 5 kg
Distance moved is 2 m

We are required to calculate the amount of work done.

How do we calculate work done?

Work done is the product of force and distance.
Therefore without distance covered no work is done.
Work done = Force × distance
Additionally, it is measured in joules or Nm

In this case;

Force = mass × 10 N/kg

= 5 kg × 10 N/kg

= 50 N

Therefore;

work done = 50 N × 2 m

= 100 Joules

Thus, the work done in lifting the object is 100 Joules

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A mass is pressed against (but is not attached to) an ideal horizontal spring on a frictionless horizontal surface. After being

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Answer:

(a) its maximum speed will be 2v and its maximum kinetic energy will be 4K.

Explanation:

The velocity of the load if it were attached would be given by

$v=\omega\sqrt{A^2-x^2}$

ω is the angular velocity of the spring, A is the amplitude and x is the displacement of the load at any time.

When the spring is compressed (or extended), then the displacement, x, is maximum and equal to the amplitude, A. When released, at its equilibrium point, x is zero. It is at this point that the unattached load moves away.

Using x = 0 in the equation,

$v=\omega A$

This is the maximum velocity. ω depends on the mass of the load and the spring constant; hence it is constant.

It follows then that $v\propto A$ .

When the compression is doubled, A is doubled. Because of the linear relationship between v and A, v is also doubled.

Kinetic energy is given by

$K = \frac{1}{2} mv^2$

When v is doubled, the kinetic energy becomes

$K_1 = \frac{1}{2} m(2v)^2 = 4(\frac{1}{2} mv^2) = 4K$

The kinetic energy is multiplied by a factor of 4.

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3 years ago

Potassium hydroxide (KOH) and hydrochloric acid (HCl) react in a beaker. They form potassium chloride (KCl) and water (H2O). Wha

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Its the second option

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4 years ago

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What is the electric field due to a point charge of 20uC at a distance of 1 meter away from it?

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The electric field due to a point charge of 20uC at a distance of 1 meter away from it is 180000 $\frac{N}{C}$ .

First, you have to know that the space surrounding a load suffers some kind of disturbance, since a load located in that space will suffer a force. The disturbance that this charge creates around it is called an electric field.

In other words, an electric field exists in a certain region of space if, when introducing a charge called witness charge or test charge, it undergoes the action of an electric force.

The electric field E created by the point charge q at any point P, located at a distance r, is defined as:

$E=K\frac{q}{r^{2} }$

where K is the constant of Coulomb's law.

In this case, you know:

K= 9×10⁹ $\frac{Nm^{2} }{C^{2} }$
q= 20 uC=20×10⁻⁶ C
r= 1 m

Replacing in the definition of electric field:

$E=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{20x10^{-6} C}{(1 m)^{2} }$

Solving:

E=180000 $\frac{N}{C}$

Finally, the electric field due to a point charge of 20uC at a distance of 1 meter away from it is 180000 $\frac{N}{C}$ .

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A standard gold bar stored at Fort Knox, Kentucky, is 7.00 inches long, 3.63 inches wide, and 1.75 inches tall. Gold has a densi

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Answer:

14.1 kg

Explanation:

Given:

Length=7.00inches

Width=3.63 inches

Height=1.75 inches

density = 19,300 kg/m3.

We can convert the given parameters to metre for unit consistency

But we know 1 inches= 0.0254 metre

✓Then Length l=7.00inches

=7×0.0254 metre=0.1778m

✓Width w =3.63 inches

==3.63 ×0.0254 metre=0.092m

✓Height h =1.75 inches

=1.75 ×0.0254 metre=0.0445 m

But Mass= density × volume

Volume= Length× width×height

Mass= density× Length× width×height

= 19300kg/m³×0.1778×0.0922×0.0445

=14.1 kg

Therefore, the mass of the gold bar is 14.1 kg

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