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melomori [17]
3 years ago
11

Vector a with arrow has a magnitude of 12.2 units and points due west. vector b with arrow points due north. (a) what is the mag

nitude of b with arrow if a with arrow + b with arrow has a magnitude of 17.4 units? 21.2 incorrect: your answer is incorrect. units (b) what is the direction of a with arrow + b with arrow relative to due west? ° correct: your answer is correct. (c) what is the magnitude of b with arrow if a with arrow â b with arrow has a magnitude of 17.4 units? units (d) what is the direction of a with arrow â b with arrow relative to due west? ° correct: your answer is correct.
Physics
1 answer:
Molodets [167]3 years ago
8 0
C = Sqrt (17.4^2 - 12.2^2) = 12.4
ArcTan (12.4/12.2) = 45.47 degrees above vector a,  which is 134.53 degrees measured on the xy-axis.

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Which part of the wave has the highest frequency?
trapecia [35]

Answer:

The last part on the right side of the diagram

Explanation:

Im on plato and just got it right :)

6 0
3 years ago
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Consider the following three objects, each of the same mass and radius: (1) a solid sphere (2) a solid disk (3) a hoop All three
Vinil7 [7]

Answer:

The correct answer is

a) 1, 2, 3

Explanation:

In rolling down an inclined plane, the potential energy is Transferred to both linear and rotational kinetic energy thus

PE = KE or mgh = 1/2×m×v² + 1/2×I×ω²

The transformation equation fom potential to kinetic energy is =

m×g×h = \frac{1}{2} mv^{2} + \frac{1}{2} (\frac{2}{5} mr^{2} )(\frac{v}{r}) ^{2}

v_{Sphere} = \sqrt{\frac{10}{7} gh}

v_{Hoop} = \sqrt{gh}

v_{Disc}=\sqrt{\frac{4}{3} gh}

Therefore the order is with increasing rotational kinetic energy hence

the first is the sphere 1 followed by the disc 2 then the hoop 3

the correct order is a, 1, 2, 3

8 0
3 years ago
3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of
Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555

Again using Snell's law for blue light as :

n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0
3 years ago
What does the zigzag line in the circuit diagram represent? A. a source of electrical energy B. an electrical conductor C. an el
lora16 [44]
C. an electrical resistor 
8 0
3 years ago
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The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g
Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:  

k=\sqrt{\frac{I}{M} }

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration  k=\sqrt{\frac{I}{M} }

So, moment of rotational inertia (I) of a cylinder about it axis = \frac{MR^2}{2}

k=\sqrt{\frac{\frac{MR^2}{2}}{M} }

k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }

k=\sqrt{{\frac{R^2}{2}}

k={\frac{R}{\sqrt{2}}

k={\frac{1.20m}{\sqrt{2}}

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = \frac{2}{3}MR^2

k = \sqrt{\frac{\frac{2}{3}MR^2}{M}  }

k = \sqrt{\frac{2}{3} R^2}

k = \sqrt{\frac{2}{3} }*R

k = \sqrt{\frac{2}{3}}  *1.20

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of  radius

(I) = \frac{2}{5}MR^2

k = \sqrt{\frac{\frac{2}{5}MR^2}{M}  }

k = \sqrt{\frac{2}{5} R^2}

k = \sqrt{\frac{2}{5} }*R

k = \sqrt{\frac{2}{5}}  *1.20

k = 0.7560

k ≅ 0.76 m

6 0
3 years ago
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