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3 years ago

5

A book is sitting on a table. Jackie is pushing the book with a force of 9 N to the right. If the force of friction is 4 N to th

e left, what is the net force?
5 N right
4 N right
4 N left
5 N left

1 answer:

Free_Kalibri [48]3 years ago

8 0

Answer:

5 N right

Explanation:

Fx = 9N-4N

Fx = 5N

Since we can define the x and y axis. We have x to the right as positive.

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An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi

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The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

$V_x'=0.8c\\V=0.6c\\m=4*10^5kg$

Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

$KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}$

So, to $V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016$ find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
We have an expression from Lorents transformation for relativistic law of addition of velocities as,

$V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V$

Substituting values, we get,

$V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c$

Thus, the KE will be,

$KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J$

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

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2 years ago

new generation cordless phones use a 9.00x10^2 MHz frequency and can be operated up to 60.0 m from their base. how many waveleng

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Λ = 3*10^8 / 9*10^8 = 1/3 m
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3 years ago

A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod

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(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
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(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
$c_6 = 0.0812m+0.1625m=0.2437 m$

(c) The total charge is $Q=-3 \cdot 10^{-8}C$ . To get the charge on each piece, we should divide this value by 8, the number of pieces:
$Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C$

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
$q= -3.75\cdot 10^{-9}C$
If we approximate piece 6 as a single charge, the electric field is given by
$E=k_e \frac{q}{d^2}$
where $k_e=8.99\cdot 10^9Nm^2C^{-2}$ and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
$E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m$
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.

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3 years ago

A single point charge is placed at the center of an imaginary cube that has 10 cm long edges. The electric flux out of one of th

12345 [234]

Answer:

Explanation:

<u>Given:</u>

Length of each side of the cube, $L=10\ cm$

The Elecric flux through one of the side of the cube is, $\phi =-1 kNm^2/C.$

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by $\epsilon_0$

Since Flux is a scalar quantity. It can added to get total flux through the surface.

$\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C$

So the the charge at the centre is calculated.

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3 years ago