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Nostrana [21]
2 years ago
9

An electrically neutral model airplane is flying in a horizontal circle on a 2.0-m guideline, which is nearly parallel to the gr

ound. The line breaks when the kinetic energy of the plane is 50 J. Reconsider the same situation, except that now there is a point charge of q on the plane and a point charge of -q at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is 53.5 J. Find the magnitude of the charges.
Physics
1 answer:
amm18122 years ago
5 0

Answer:

q=3.5*10^-4

Explanation:

<u>concept:</u>

The force acting on both charges is given by the coulomb law:

F=kq1q2/r^2

the centripetal force is given by:

Fc=mv^2/r

The kinetic energy is given by:

KE=1/2mv^2

<u>The tension force:</u>

<u><em>when the plane is uncharged </em></u>

T=mv^2/r

T=2(K.E)/r

T=2(50 J)/r

T=100/r

<u><em>when the plane is charged </em></u>

T+k*|q|^2/r^2=2(K.E)charged/r

100/r+k*|q|^2/r^2=2(53.5 J)/r

q=√(2r[53.5 J-50 J]/k)                                          √= square root on whole

q=√2(2)(53.5 J-50 J)/8.99*10^9

q=3.5*10^-4

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kherson [118]
The appropriate response is letter D. The wave ventures slower and with an expanded wavelength when a sound wave entering a range of hotter air. Hotter air implies less thick, so the wave ought to back off.

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2 years ago
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A 2.0 kg block has a rope attached to the block on a table and is pulled with a force of 8.0 N. The block accelerated at 2.5m/s^
Masja [62]

Answer:

0.15

Explanation:

Assuming the rope is horizontal, sum the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum the forces in the x direction:

∑F = ma

F − Nμ = ma

Substitute:

F − mgμ = ma

mgμ = F − ma

μ = (F − ma) / (mg)

Plug in values:

μ = (8.0 N − 2.0 kg × 2.5 m/s²) / (2.0 kg × 9.8 m/s²)

μ = 0.15

3 0
2 years ago
The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Dete
jeka57 [31]

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

y=\frac{0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m}{(\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m} \\\\y=0.88\ m

b)  the mass moment of inertia about z axis passing the rotation center O is:

I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2

c) The mass moment of inertia about z axis passing the rotation center O is:

I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2

3 0
3 years ago
In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de
snow_tiger [21]

As we know that range of the projectile motion is given by

R = \frac{v^2 sin(2\theta)}{g}

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

\theta, 90 - \theta

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

5 0
2 years ago
Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the
tatuchka [14]

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

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a = Acceleration due to gravity Earth= 9.81 m/s²

Accelration going up is considered as negetive

Initial Velocity of the ball

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m

The displacement of the ball on the moon is 7.04 m

6 0
3 years ago
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