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3 years ago

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A test charge is placed at a distance of 3.5 meters from a charge of 8.7 × 10-9 coulombs. What is the electric field strength at

the test charge? (k = 9.0 × 109 newton·meter2/coulomb2) 6.4 newtons/coulomb 8.7 newtons/coulomb 3.5 newtons/coulomb 9.0 newtons/coulomb Done

1 answer:

zavuch27 [327]3 years ago

8 0

In order to calculate the electric field strength, we may use the formula:
E = kQ/d²
Where Q is the charge and d is the distance between the charge and the test charge. Substituting the values into the equation:
E = (9 x 10⁹)(8.7 x 10⁻⁹) / (3.5²)
E = 6.39 Newtons per coulomb

Therefore, the answer is 6.4 Newtons/coulomb

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0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for

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Assuming the accleration applied was constant, we have

$v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)$

$\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)$

$\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}$

Then the force applied to the ball is given by

$F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)$

$\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N$

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3 years ago

A balloon filled with helium gas has an average density of Q,-0.41 kg/m'. The density of the air is Qa-1.23 kg/m3. The volume of

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Answer:

a) $(Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]$

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Explanation:

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$massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\$

The buoyancy force acting on the balloon is:

$Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]$

Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.

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3 years ago

For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equat

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Answer:

Radius of the solenoid is 0.93 meters.

Explanation:

It is given that,

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$B(t)=5t\ T$ , t is time in seconds

$\dfrac{dB}{dt}=5\ T$

The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m

The electric field due to changing magnetic field is given by :

$E(2\pi x)=\dfrac{d\phi}{dt}$

x is the distance from the axis of the solenoid

$E(2\pi x)=\pi r^2\dfrac{dB}{dt}$ , r is the radius of the solenoid

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So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.

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Read 2 more answers

Prove(show) ''T=2π√(l/g)''

Nonamiya [84]

Answer:

Time period for Simple pendulum, $T=2\pi\sqrt{\frac{l}{g}$

Explanation:

The Simple Pendulum

Consider a small bob of mass $m$ is tied to extensible string of length $l$ that is fixed to rigid support. The bob is oscillating in the plane about verticle.

Let $\theta$ is the angle made by string with vertical during oscillation.

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Negative sign shows that its opposing the motion of bob.

Taking $\theta$ as very small angle then, $\sin\theta\sim\theta$

$F=-mg\theta$

Let $x$ is the displacement made by bob from its mean position ,

then, $\theta=\frac{x}{l}$

so, $F=-mg\frac{x}{l}$ ........(1)

Since, pendulum is in hormonic motion,

as we know, $F=-kx$

where $k$ is the constant and $k=m\omega^{2}$

$F=-m\omega^2x$ .........(2)

From equation (1) and (2)

$-m\omega^2x=-mg\frac{x}{l}$

$\omega=\sqrt{\frac{g}{l}}$

Since, $\omega=\frac{2\pi}{T}$

$\frac{2\pi}{T}=\sqrt{\frac{g}{l}$

$T=2\pi\sqrt{\frac{l}{g}}$

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DerKrebs [107]

Answer:

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For uniform velocity, the change in velocity, Δv = 0

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3 years ago