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alexdok [17]
3 years ago
8

The magnitude of displacements a and b are 3m and 4m, respectively, c=a+b. What is the magnitude of c if the angel between a and

b is (a) 0 and (b) is 180?
Physics
1 answer:
AysviL [449]3 years ago
4 0

Answer:

(a) 7 m

(b) 1  m

Explanation:

Given:

The magnitude of displacement  vector 'a' is 3 m

The magnitude of displacement vector 'b' is 4 m.

The vector 'c' is the vector sum of vectors 'a' and 'b'.

(a)

Now, when the angle between the vectors is 0°, it means that the vectors are in the same direction. When vectors are in the same direction, then their resultant magnitude is simply the sum of their magnitudes.

So, magnitude of 'c' when 'a' and 'b' are in same direction is given as:

|\overrightarrow c|=|\overrightarrow a|+|\overrightarrow b|\\\\|\overrightarrow c|=3 + 4 = 7\ m

Therefore, the magnitude of vector 'c' is 7 m when angle between 'a' and 'b' is 0°.

(b)

When the angle between the vectors is 180°, it means that the vectors are exactly in the opposite direction. When the vectors are in opposite direction, then their resultant magnitude is the subtraction of their magnitudes.

So, magnitude of 'c' when 'a' and 'b' are in opposite direction is:

|\overrightarrow c|=||\overrightarrow a|-|\overrightarrow b||\\\\|\overrightarrow c|=|3 - 4| = 1\ m

Therefore, the magnitude of vector 'c' is 1 m when angle between 'a' and 'b' is 180°.

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Which of the following is true about all of the outer planets?A.Much of the material in these planets is solid.B.The surface of
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3 years ago
Determine the minimum size of glass tubing that can be used to measure water level. If the capillary rise in the tube does not e
Marina CMI [18]

Answer:

Explanation:

The formula to determine the size of a capillary tube is

h = 2•T•Cos θ / r•ρ•g

Where

h = height of liquid level

T = surface tension

r = radius of capillary tube

ρ = density of liquid

θ = angle of contact = 0°

g =acceleration due to gravity=9.81m/s²

The liquid is water then,

ρ = 1000 kg / m³

Given that,

T = 0.0735 N/m

h = 0.25mm = 0.25 × 10^-3m

Then,

r = 2•T•Cos θ / h•ρ•g

r = 2 × 0.0735 × Cos0 / 2.5 × 10^-3 × 1000 × 9.81

r = 5.99 × 10^-3m

Then, r ≈ 6mm

The radius of the capillary tube is 6mm

So, the minimum size is

Volume = πr²h

Volume = π × 6² × 0.25

V = 2.83 mm³

The minimum size of the capillary tube is 2.83mm³

8 0
3 years ago
A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl
kondaur [170]

Answer:

Part a)

y_m = 0.157 mm

part b)

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

Explanation:

As we know that the speed of wave in string is given by

v = \sqrt{\frac{T}{m/L}}

so we have

T = 17.5 N

m/L = 5.4 g/cm = 0.54 kg/m

now we have

v = \sqrt{\frac{17.5}{0.54}}

v = 5.69 m/s

now we have

Part a)

y_m = amplitude of wave

y_m = 0.157 mm

part b)

k = \frac{\omega}{v}

here we know that

\omega = 2\pi f

\omega = 2\pi(92.2) = 579.3 rad/s

so we  have

k = \frac{579.3}{5.69}

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

4 0
3 years ago
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