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Salsk061 [2.6K]
4 years ago
5

Vector 1 points along the z axis and has magnitude V1 = 80. Vector 2 lies in the xz plane, has magnitude V2 = 50, and makes a -3

7° angle with the x axis (points below the x axis). What is the scalar product V1·V2?
Physics
1 answer:
Y_Kistochka [10]4 years ago
5 0

Answer:

\vec{V1} . \vec{V2}  = 2574.08    

Explanation:

given data

magnitude V1 = 80

magnitude V2 = 50

angle a =  -37°

solution

\vec{V1} = 80 k

\vec{V2} = 50 cos{37} i  - 50sin{37} k

so that here \vec{V1} . \vec{V2}    is

\vec{V1} . \vec{V2}  = 80 k . ( 50 cos{37} i  - 50sin{37} k )

\vec{V1} . \vec{V2}  = 80 k . (  38.270 i + 32.176 k )

\vec{V1} . \vec{V2}  = 2574.08    

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4 0
3 years ago
PART 2 OF ENERGY AND FORCES UNIT TEST
katrin2010 [14]

Answer:

1. at least two charged interacting parts

2. from the electric fields of charged subatomic particles

3 an arrow released from the bow

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8 Iron pieces accelerate toward the magnet, and the energy stored in the system decreases.

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4 0
3 years ago
To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a
irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

                                                                                                               = 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

                                           W = F x S    joules

Substituting the values in the above equation

                                            W = 1.5 x 10⁶  x  355 · sin θ

                                                = 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J                        

6 0
3 years ago
A 15-µF capacitor and a 25-µF capacitor are connected in parallel, and charged to a potential difference of 60 V. How much energ
Alborosie

Answer:

Energy stored, E = 0.072 J

Explanation:

Given that,

Capacitance, C_1=15\ \mu F

Capacitance, C_2=25\ \mu F

These two capacitor are connected in parallel, and charged to a potential difference of, V = 60 volts

We know that in parallel combination of capacitor, the equivalent capacitance is given by :

C=C_1+C_2\\\\C=(15+25)\ \mu F\\\\C=40\times 10^{-6}\ F

The energy stored in the capacitor is given by :

E=\dfrac{1}{2}CV^2\\\\E=\dfrac{1}{2}\times 40\times 10^{-6}\times (60)^2\\\\E=0.072\ J

So, the energy stored in the capacitor in this capacitor combination is 0.072 J.

4 0
4 years ago
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