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3 years ago

Potential energy can be described as situations?

Physics

1 answer:

hodyreva [135]3 years ago

6 0

Potential energy is the energy of an object stored when it is kept from a high distance

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State a Newtowns second law

astraxan [27]

Answer:

the second law states that the force F is the product of an object's mass and its acceleration a: F = m * a. For an external applied force, the change in velocity depends on the mass of the object.

5 0

3 years ago

A 5.00 kg block of metal with

Nitella [24]

Answer:

953.5 J/kg.°C

Explanation:

From the question,

Heat lost by the metal = heat gained by the glass.

cm(t₁-t₃) = c'm'(t₃-t₂)................. Equation 1

Where c = specific heat capacity of the metal, m = mass of the metal, c' = specific heat capacity of the glass, m' = mass of the glass, t₁ = initial temperature of metal, t₂ = initial temperature of glass, t₃ = Equilibrium temperature

Make c' the subject of the equation

c' = cm(t₁-t₃)/m'(t₃-t₂)................ Equation 2

Given: m = 5 kg, c = 650 J/kg.°C, m' = 1.25 kg, t₁ = 80 °C, t₂ = 20 °C, t₃ = 63.9 °C

Substitute these values into equation 2

c' = 5×650(80-63.9)/1.25(63.9-20)

c' = (5×650×16.1)/(1.25×43.9)

c' = 52325/54.875

c' = 953.5 J/kg.°C

5 0

3 years ago

Read 2 more answers

A farmer places unhatched chicken eggs under a heat lamp. How does the radiation help the eggs?

SOVA2 [1]

Radiation helps chicken eggs warm. (The eggs need to be kept at a warm temperature so the chicks can be born naturally and safely).

7 0

4 years ago

Read 2 more answers

There are two balloons of charges +3.37 x 10-6 C and –8.21 x 10-6 C. The distance between the two balloons is 2.00 m. Determine

andrew-mc [135]

F = 0.06N. Since the charges has different signs the force of atraction between them is 0.06N.

In order to solve this exercise we have to use Coulomb's Law equation which says that the magnitude of each of the electric forces with which two point charges at rest interact is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them. Given by the equation:

$F = k\frac{|q_{1}q_{2}|}{d^{2}}$ . Where k is the Coulomb's Constant $k = 9x10^{9}\frac{Nm^{2} }{C^{2} }$ , q1 and q2 are the charges value in Coulomb (C), and d is the distance between charges in meters (m).

There are two balloons of charges +3.37 x 10-6 C and –8.21 x 10-6 C. The distance between the two balloons is 2.00 m. Calculate the force between the two ballons.

$F = 9x10^{9}\frac{Nm^{2} }{C^{2} } \frac{|(3.37x10^{-6}C)(-8.21x10^{-6}C)|}{(2.00m)^{2}}\\F = 9x10^{9}\frac{Nm^{2} }{C^{2} } \frac{|-2.77x10^{-11}C^{2} )|}{4.00m^{2}}\\F = 9x10^{9}\frac{Nm^{2} }{C^{2} }(6.92x10^{-12})\frac{C^{2} }{m^{2}}\\F = 0.06N$

4 0

3 years ago

An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert

cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, $k=4.2\times 10^5\ N/m$

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

$mgh=\dfrac{1}{2}kx^2$

x is the compression in spring

$x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m$

So, the compression in the spring is 5.88 meters.

6 0

3 years ago