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son4ous
2 months ago
11

The image shows the displacement of a motorboat. The data table shows the magnitudes of the components of each displacement vect

or.
What is the value of Rx?

km

What is the value of Ry?

km

Physics
2 answers:
serg [3.5K]2 months ago
8 0

R_x = A_x + B_x + C_x

A_x is negative since the vector resides in the third quadrant.

B_x is positive as it is directed in the first quadrant.

C_x is positive because it points towards the fourth quadrant.

Thus,

R_x = -1.6 + 3.9 + 1.2

R_x = 3.5 km

R_y = A_y + B_y + C_y

A_y is negative as it exists in the third quadrant.

B_y is positive because it's in the first quadrant.

C_y is negative since it is in the fourth quadrant.

R_y = -2.8 + 6.4 - 0.7

R_y = 2.9 km

serg [3.5K]2 months ago
4 0
Rx= 3.5 km

Ry= 2.9 km
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6 0
1 month ago
A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac
Keith_Richards [3271]

Answer:

The flux across the cube's surface is 2.314\ Nm^{2}/C.

Solution:

According to the details provided:

Cube edge length, a = 8.0 cm = 8.0\times 10^{- 2}\ m.

Volume charge density, \rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}.

Now,

To find the electric flux:

\phi = \frac{q}{\epsilon_{o}}

where

\phi = electric flux

\epsilon_{o} = 8.85\times 10^{- 12}\ F/m = permittivity of vacuum.

The volume charge density for this scenario is described by:

\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}

Cube volume, V = a^{3}.

Thus,

V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}.

The total charge can be derived from equation (2):

q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}.

q = 2.048\times 10^{-11}\ F = 20.48\ pF.

Now, insert the value of 'q' into equation (1):

\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C.

5 0
2 months ago
To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Softa [3030]

Answer:

H = 10.05 m

Explanation:

The stone reaches the top of the flagpole at both t = 0.5 s and t = 4.1 s

therefore, the total duration of the upwards motion above the peak of the pole is provided as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

this indicates the speed at the flagpole's top

at this point we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

the height of the flagpole is stated as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
3 months ago
Two charges q1 = 5 µC, q2 = -26 µC, are L = 19 cm apart. A third charge is to be placed on the line between the two charges. How
Keith_Richards [3271]
The electric force between two objects is expressed as being proportional to the product of their charges and inversely proportional to the square of the distance separating them. In this instance, the distance between the first two charges is 19 cm. We formulate the equation k q1 q3/ (x)^2 = k q2 q3/ (19-x)^2, where x denotes the separation between q1 and q3. The charge q3 cancels out, and q2 is used in absolute terms. The resulting value of x is 5.79 cm.
6 0
1 month ago
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