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adelina 88 [10]
3 years ago
5

Various radial points on a rotating Ferris wheel have: I. different linear velocities II. different angular velocities III. equa

l linear velocities IV. equal angular velocities
Physics
1 answer:
Oliga [24]3 years ago
4 0

Answer:

I) True,  II) False,  III) False,  IV) True    

Explanation:

In this exercise, it is asked to answer different statements, for this we will use the relationship between angular and linear velocity

           v = w r

let's review the claims

I) True. From the initial equation we see that the linear velocity depends on the radius

II) False. All points rotate with the same angular velocity

III) False. Linear velocity changes with radius

IV) True. The angular velocity of all points is the same

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Express the vector R<br> B<br> in terms of A, B, C, and Ď, the edges of a<br> parallelogram.
Vadim26 [7]

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;

7 0
3 years ago
If the diameter of a radar dish is doubled, what happens to its resolving power assuming that all other factors remain unchanged
kirill115 [55]

Answer:

      θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

Explanation:

The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.

The first minimum occurs for m = 1, so the diffraction equation of a slit remains

        a sin θ = λ

in general, the diffraction patterns occur at very small angles, so

        sin θ = θ

          θ = λ / a

in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.

        θ = 1.22 λ /a

In this exercise we are told that the opening changes

         a’ = 2 a

we substitute

          θ ‘= 1.22  λ / 2a

          θ' = (1.22 λ / a) 1/2

          θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

8 0
3 years ago
You are looking at a yellow flower growing outside in the sunshine. Why does it look yellow?
Alchen [17]

Answer:

It looks yellow because that is the only (major) color reflected.

Visible spectra is from about 4000-7000 Angstroms (10^-10 m).

Red are longer wavelengths and blue are the shorter wavelengths.

The Sodium doublet (yellow) occurs around 5900 Angstroms.

3 0
2 years ago
A particle leaves the origin with an initial velocity v → = (3.00iˆ) m/s and a constant acceleration a → = (−1.00iˆ − 0.500jˆ) m
tatiyna

Answer:

the position vector (x,y) will be (1.5 m,-2.25 m) and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s) when x reaches its maximum x coordinate

Explanation:

Since the velocity is related with the acceleration and coordinates through

vx²=v₀x²+2*ax*x

where

vx = velocity in the x direction

v₀x = initial velocity in the x direction = 3 m/s

ax = acceleration in the x direction = −1.00 m/s²

x= coordinates in the x-axis

when x reaches its maximum coordinate , then vx=0

thus

vx²=v₀x²+2*ax*x

0 = (3 m/s)² + 2* (−1.00 m/s²)*x

x= 1.5 m

also for the time t

vx = v₀x + ax*t → t= (vx-v₀x)/ax = (0- 3 m/s)/  (−1.00 m/s²) = 3 seconds

for the y coordinates

y = y₀+v₀y*t + 1/2 ay*t²

where

v₀y = initial velocity in the y direction = 0 m/s

ay = acceleration in the x direction = −0.5 m/s²

y= coordinates in the y-axis

y₀= initial coordinate in the y-axis =0

then since y₀=0 and v₀y=0

y = 1/2*ay*t²

y = 1/2*ay*t² = 1/2*(−0.5 m/s²)*(3 s)² = -2.25 m

and

vy=v₀y+ ay*t= 0+(−0.5 m/s²)*(3 s)= (-1.5 m/s)

therefore the position vector (x,y) will be (1.5 m,-2.25 m)

and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s)

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4 years ago
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