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3 years ago

What is a non-example of gravitational force?

2 answers:

6 0

The force that holds two magnets together is the ONLY non-example
of gravitational force. Except for the strong nuclear force, the weak
nuclear force, the electrostatic force, the tidal force, the US Air Force,
the force of destiny, the force of faith, the strength of one's convictions,
and a few million other non-examples.

qaws [65]3 years ago

4 0

A non-example of force would be something that stay sill like a balloon in the air..... taste, smell, feel, texture, color, opinion, faith, hope, sincerity, honest, speed, momentum, altitude, volume, loudness, area, length, acidity, obesity, nationalism, current, resistance, viscosity, wavelength, flow, rate, frequency, albedo, diameter, age, temperature, acceleration, body mass index, salinity, specific, specific gravity, consciousness, intelligence, refraction index, mass, time, date rate, switching speed, libido, focal length, and latency are not force. And even there are many other things that also are not force, too.

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A 0.130 m radius, 485-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a

Lady_Fox [76]

Answer:

The magnetic field strength needed is 1.619 T

Explanation:

Given;

Number of turns, N = 485-turn

Radius of coil, r = 0.130 m

time of revolution, t = 4.17 ms = 0.00417 s

average induced emf, V = 10,000 V.

Average induced emf is given as;

V = -ΔФ/Δt

where;

ΔФ is change in flux

Δt is change in time

ΔФ $= -NBA(Cos \theta_f - Cos \theta_i)$

where;

N is the number of turns

B is the magnetic field strength

A is the area of the coil = πr²

θ is the angle of inclination of the coil and the magnetic field,

$\theta_f = 90^o\\\theta_i = 0^o$

V = NBACos0/t

V = NBA/t

B = (Vt)/NA

B = (10,000 x 0.00417) / (485 x π x 0.13²)

B =1.619 T

Thus, the magnetic field strength needed is 1.619 T

5 0

3 years ago

Read 2 more answers

PLS HELP BEING TIMED!!!

Elodia [21]

Answer:

$K_{i}+U_{g,i} = K_{f}+U_{g,f}$

Explanation:

A closed system is a system where exists energy interactions with surroundings, but not mass interactions. If we neglect any energy interactions from boundary work, heat, electricity, magnetism and nuclear phenomena and assume that process occurs at steady state and all effects from non-conservative forces can be neglected, then the equation of energy conservation is reduce to this form:

$\Delta K +\Delta U_{g} = 0$ (1)

Where:

$\Delta K$ - Change in kinetic energy of the system, measured in joules.

$\Delta U_{g}$ - Change in gravitational potential energy of the system, measured in joules.

If we know that $\Delta K=K_{i}-K_{f}$ and $\Delta U_{g} = U_{g,i}-U_{g,f}$ , then we get the following equation:

$K_{i}+U_{g,i} = K_{f}+U_{g,f}$ (2)

Where $i$ and $f$ stands for initial and final states of each energy component.

Hence, the right answer is $K_{i}+U_{g,i} = K_{f}+U_{g,f}$

7 0

3 years ago

Read 2 more answers

How do we know Earth is spinning, not the Sun?

Tpy6a [65]

Answer:

Because it is said that the earth rotates and revolves around the sun and moon so it is impossible for the earth not to be spinning.

Explanation:

4 0

3 years ago

A driver, traveling at 22.0 m/s, slows down her 2.00 x 103 kg car to stop for a red light. What work is done by the friction for

Artemon [7]

Some work will be done on friction between wheels and road but it is negligible compared to work done on friction on breaks.

W = Ek = (m*v^2)/2 = 2000*22^2/2 = 1000*22^2 = 484KJ

Because car is not changing its potential energy, there is no work to be done on while changing it which means that all goes on changing kinetic energy (energy of motion)

6 0

3 years ago

Kyle, a 95.0 kg football player, leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.

babymother [125]

Answer:

The initial speed of the ball was 26.2 m/s

Explanation:

When the football player is in the air at his maximum height the vertical component of velocity is zero, To obtain the horizontal velocity when the player catches the ball we need to apply the linear momentum conservation theorem:

$m_1*v_{o1}+m_2*v_{o2}=m_t*v_f\\0.430kg*(v)+m_2*(0)=mt*vt$

we need to obtain the time taken to go down.

$y=Y_o+v_o*t-\frac{1}{2}g*t^2\\\\0=0.589-4.9t^2\\solving:\\t_1=0.346s\\$

We have a horizontal displacement and the time taken to stop, so:

$v_f=\frac{d}{t}=\frac{0.0409m}{0.346s}=0.118m/s$

so:

$0.430kg*(v)+m_2*(0)=(m1+m2)*vt\\v=\frac{(95.0kg+0.430kg)*0.118m/s}{0.430kg}\\\\v=26.2m/s$

8 0

2 years ago