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2 years ago

13

If vx=9.80 units and vy=-6.40 units, determine the magnitude and direction of v

1 answer:

dexar [7]2 years ago

5 0

The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:

v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units

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A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. (a) What is the foca

xenn [34]

Answer:

a) Focal length of the lens is 8 cm which is a convex lens

b) 6 cm

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

u = Object distance = 4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm$

a) Focal length of the lens is 8 cm which is a convex lens

Magnification

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2$

b) Height of image is 2×3 = 6 cm

Since magnification is positive the image upright

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

8 0

2 years ago

A source charge generates an electric field of 1236 N/C at a distance of 4 m. What is the magnitude of the source charge?

guajiro [1.7K]

The correct answer to the question is- $2.2\ \mu C$

CALCULATION:

As per the question, the electric field generated by the source charge is 1236 N/C at a distance of 4 m.

Hence , electric field E = 1236 N/C.

The distance of the point R = 4m

We are asked to calculate the charge possessed by the source.

The electric field produced by a source charge of Q at a distance R is calculated as -

Electric field E = $\frac{1}{4\pi \epsilon_{0}}\frac{Q}{R^2}$

Here, $\epsilon_{0}$ is called the absolute permittivity of the free space.

Hence, the charge of source is calculated as -

Q = $E\times 4\pi \epsilon_{0}\times R^2$

= $1236\times \frac{1}{9\times 10^9}\times (4)^2\ Coulomb$

= $2197.33\times 10^{-9}\ C$

= $2.19733\times 10^{-6}\ C$

= $2.2\ \mu C$

Hence, the charge of source is $2.2\ \mu C$

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2 years ago

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deff fn [24]

Answer:

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2 years ago

The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end

Xelga [282]

Answer:

Explanation:

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Thus there is one additional resonance corresponding to n=1 , between $f_1$ and $f_2$

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