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2 years ago

Define the term potential energy of a body? How is it measured

Physics

2 answers:

kondaur [170]2 years ago

7 0

Answer:

The energy possessed by a body as a result of its position or condition rather than its motion.

Simplified, this formula can be written as: Potential Energy = mgh, where m is the mass, measured in kilograms; g is the acceleration due to gravity (9.8 m/s^2 at the surface of the Earth); and h is the height, measured in meters.

DENIUS [597]2 years ago

4 0

Answer:

potential energy. noun. the energy of a body or system as a result of its position in an electric, magnetic, or gravitational field. It is measured in joules (SI units), electronvolts, ergs, etcSymbol: E p, V, U, φ Abbreviation: PE.

Explanation:

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1 point

e-lub [12.9K]

The engineer built a device called a generator

4 0

3 years ago

Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is

Llana [10]

Answer:

$K_e_q=22.75878093\frac{N}{m}$

$f=1.363684118Hz$

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

$\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}$

Replacing the data provided:

$\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}$

$K_e_q=22.75878093\frac{N}{m}$

Finally, to calculate the frequency of oscillation we use this:

$f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }$

Replacing m and k:

$f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz$

4 0

3 years ago

Un automovil parte del reposo y acelera uniformemente hasta alcanzar una rapidez de 0,255km/h en un tiempo de 3/4 Minutos determ

Elden [556K]

Answer:

a = 1.5*10^-3 m/s^2

x = 0.033m = 3.3cm

Explanation:

To calculate the acceleration and the distance traveled by the car you use the following formulas:

$v=v_o+at$ (1)

$x=v_ot+\frac{1}{2}at^2$ (2)

v: final velocity = 0,255 km/h

vo: initial velocity = 0 m/s

t: time = 3/4 min

a: acceleration = ?

x: distance

In order to use the equations (1) and (2) you first convert the units of the final velocity to m/s, and the time to seconds.

$v=0,255\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}\\\\v=0.07m/s\\\\t=\frac{3}{4}min*\frac{60s}{1min}=45s$

Next, you solve the equation (1) for the acceleration a:

$a=\frac{v}{t}=\frac{0.07m/s}{45s}=1.5*10^{-3}\frac{m}{s^2}$

With this value of a you can calculate the distance traveled by the car, by using the equation (2):

$x=\frac{1}{2}(1.5*10^{-3}m/s^2)(45s)^2=0.033m=3.3cm$

hence, the acceleration of the car is 1.5*10^-3 m/s^2 and the distance traveled in 3/4 min is 0.033m

5 0

3 years ago

A candle is 49 cm in front of a convex spherical mirror that has a focal length

vitfil [10]

The convex spherical mirror is the diverging mirror. The image distance is 20.4 cm. The magnification is 0.417. The image is virtual and upright.

<h3>What is magnification?</h3>

Magnification is the ratio of image distance to the object distance.

Object distance is 49 cm and the focal length of convex mirror is 35 cm, then image distance is calculated by the lens maker formula.

1/f = 1/v +1/u

1/-35 = 1`/v +1/49

v= -20.4 cm

Thus, the image distance is 20.4 cm.

The magnification is given by

m = v/u

m = -20.4 / 49

m = 0.417

Thus, the magnification is 0.417.

The image formed is by virtual meeting of light rays and above the principle axis. Thus the image is virtual and upright.

Learn more about magnification.

brainly.com/question/21370207

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5 0

2 years ago

Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis. The magnitude of q1 is 3 times the magnitu

Alex777 [14]

d = distance between the two point charges = 60 cm = 0.60 m

r = distance of the location of point "a" where the electric field is zero from charge $q_{1}$ between the two charges.

$q_{1}$ = magnitude of charge on one charge

$q_{2}$ = magnitude of charge on other charge

$q_{1}$ = 3 $q_{2}$

$E_{1}$ = Electric field by charge $q_{1}$ at point "a"

$E_{2}$ = Electric field by charge $q_{2}$ at point "a"

Electric field by charge $q_{1}$ at point "a" is given as

$E_{1}$ = k $q_{1}$ /r²

Electric field by charge $q_{2}$ at point "a" is given as

$E_{2}$ = k $q_{2}$ /(d-r)²

For the electric field to be zero at point "a"

$E_{2}$ = $E_{1}$

k $q_{2}$ /(d-r)² = k $q_{1}$ /r²

$q_{2}$ /(d-r)² = 3 $q_{2}$ /r²

1/(0.60 - r)² = 3 /r²

r = 0.38 m

r = 38 cm

8 0

3 years ago