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3 years ago

Express the vector R B in terms of A, B, C, and Ď, the edges of a parallelogram.

Physics

1 answer:

Vadim26 [7]3 years ago

7 0

Answer:

R=0.5B+0.5C+2A+D

Explanation:

By the triangular law of vector addition

vector R= vector B- vector D

As A,B,C,D are edges of the parallelogram,

A is parallel to D but opposite in direction.

Therefore

$A = (-D)$ ; $A//D$ ;

$2A=-2D$

B is parallel to C and in same direction.

$B//C\\B=C\\$

$0.5B=0.5C\\$

$R= B-D;\\R= 0.5B+0.5B-D;\\R=0.5B+0.5C-D;\\R=0.5B+0.5C-2D+D;\\R=0.5B+0.5C+2A+D;$

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S Problem Set<br> 2.) 6.4 x 109 nm to cm

anyanavicka [17]

Answer:

$6.4\cdot 10^2 cm$

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

$1 nm = 10^{-9} m$

So we have:

$6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m$

Now we can convert from metres to centimetres, keeping in mind that

$1 m = 10^2 cm$

So, we find:

$6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm$

8 0

3 years ago

The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an

Readme [11.4K]

1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
$d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
substituting the coordinates of the two charges, we get
$d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m$

2) Then, we can calculate the electrostatic force between the two charges $q_1$ and $q_2$ , which is given by
$F=k_e \frac{q_1 q_2}{d^2}$
where $k_e=8.99\cdot10^{9} Nm^2C^{-2}$ is the Coulomb's constant.
Substituting numbers, we get
$F=8.99\cdot10^{9} Nm^2C^{-2} \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N$
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.

7 0

4 years ago

Un cubo de madera de densidad 0.780 g/cm³ mide 11.2 cm en un lado. Cuando se coloca en agua, ¿qué altura del bloque flotará sobr

Stolb23 [73]

Answer:

2.464 cm above the water surface

Explanation:

Recall that for the cube to float, means that the volume of water displaced weights the same as the weight of the block.

We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3 cm^3):

weight of the block = 0.78 * 11.2^3 gr

Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.

So the weight of the volume of water displaced is:

weight of water = 1 * 11.2^2 * x

we make both weight expressions equal each other for the floating requirement:

0.78 * 11.2^3 = 11.2^2 * x

then x = 0.78 * 11.2 cm = 8.736 cm

This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:

11.2 cm - 8.736 cm = 2.464 cm

6 0

3 years ago

Which option is the most accurate description of speed?

puteri [66]

B I think is the answer

5 0

3 years ago

A floating ice block is pushed through a displacement along a straight embankment by rushing water, which exerts a force on the

lions [1.4K]

The question is incomplete. Here is the complete question.

A floating ice block is pushed through a displacement vector d = (15m)i - (12m)j along a straight embankment by rushing water, which exerts a force vector F = (210N)i - (150N)j on the block. How much work does the force do on the block during displacement?

Answer: W = 4950J

Explanation: <u>Work</u> (W), in physics, is done when a force acts on an object that has a displacement form a place to another:

W = F · d

As the formula shows, Work is a scalar product, i.e, it results in a number, so, Work only has magnitude.

Force and displacement for the ice block are in 2 dimensions, then work will be:

W = (210)i - (150)j · (15)i - (12)j

W = (210*15) + (150*12)

W = 3150 + 1800

W = 4950J

During the displacement, the ice block has a work of 4950J

8 0

3 years ago