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ankoles [38]
3 years ago
6

Why do all objects above earth's surface have gravitational potential energy

Physics
1 answer:
gayaneshka [121]3 years ago
6 0
Gravitational potential energy<span> is </span>energy<span> an object possesses because of its position in a </span>gravitational<span> field. The most common use of </span>gravitational potential energy<span> is for an object near the surface of the Earth where the </span>gravitational<span> acceleration can be assumed to be constant at about 9.8 m/s</span>2<span>.</span>
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What are the kinetic energy​
bearhunter [10]

Answer:

n physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion.[1] It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.

In classical mechanics, the kinetic energy of a non-rotating object of mass m traveling at a speed v is {\displaystyle {\begin{smallmatrix}{\frac {1}{2}}mv^{2}\end{smallmatrix}}}{\begin{smallmatrix}{\frac {1}{2}}mv^{2}\end{smallmatrix}}. In relativistic mechanics, this is a good approximation only when v is much less than the speed of light.

The standard unit of kinetic energy is the joule, while the imperial unit of kinetic energy is the foot-pound.

Explanation:

6 0
3 years ago
Read 2 more answers
Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example
iragen [17]

Given Information:

Magnetic field = B = 1×10⁻³ T

Frequency = f = 72.5 Hz

Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m

Required Information:

Maximum Emf = ?

Answer:

Maximum Emf = 20.66×10⁻¹² volts

Explanation:

The maximum emf generated around the perimeter of a cell in a field is given by

Emf = BAωcos(ωt)

Where A is the area, B is the magnetic field and ω is frequency in rad/sec

For maximum emf cos(ωt) = 1

Emf = BAω

Area is given by

A = πr²

A = π(d/2)²

A = π(7.60×10⁻⁶/2)²

A = 45.36×10⁻¹² m²

We know that,

ω = 2πf

ω = 2π(72.5)

ω = 455.53 rad/sec

Finally, the emf is,

Emf = BAω

Emf = 1×10⁻³*45.36×10⁻¹²*455.53

Emf = 20.66×10⁻¹² volts

Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts

8 0
3 years ago
You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s
densk [106]

Here when car in front of us applied brakes then it is slowing down due to frictional force on it

So here we can say that friction force on the car front of our car is given as

F_f = \mu m g

So the acceleration of car due to friction is given as

F_{net} = - \mu mg

a = \frac{F_{net}}{m}

a = -\mu g

now it is given that

\mu = 0.868

g = 9.81 m/s^2

so here we have

a = -0.868 * 9.81

a = -8.52 m/s^2

so the car will accelerate due to brakes by a = - 8.52 m/s^2

4 0
3 years ago
Which measurements are equal to 321 decimeters? Check all that apply.
S_A_V [24]
32,100 Millimeters
3,210 Centimeters
3.21 Decameters

Hope It Helps
5 0
3 years ago
Read 2 more answers
Help with the two questions above? Correct answers?
LenKa [72]

(6) first choice: the frequency appears higher and wavelength is shorter.

The car approaches a stationary observer and so the sound will appear to have shorter wavelength. This creates an effect of its siren to sound with higher frequency than it would do if both were stationary.

(7) The Doppler formula for frequency in the case of a stationary observer and source approaching it is as follows:

f_O = \frac{v}{v-v_s}\cdot f= \frac{343\frac{m}{s}}{(343-25)\frac{m}{s}}\cdot 400Hz \approx 431Hz

The wavelength is then

\lambda = \frac{343\frac{m}{s}}{431Hz}\approx 0.80 m

The third choice "0.80m; 431Hz" is correct

7 0
3 years ago
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