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Sedbober [7]
3 years ago
11

The image shows the displacement of a motorboat. The data table shows the magnitudes of the components of each displacement vect

or.
What is the value of Rx?

km

What is the value of Ry?

km

Physics
2 answers:
Neporo4naja [7]3 years ago
8 0

R_x = A_x + B_x + C_x

A_x is negative as the vector is in third quadrant.

B_x is positive as it points in the first quadrant.

C_x is positive as it points in the fourth quadrant.

Hence,

R_x = -1.6 + 3.9 + 1.2

R_x = 3.5 km

R_y = A_y + B_y + C_y

A_y is negative as it is in third quadrant.

B_y is positive as it is in first quadrant.

C_y is negative as it is in fourth quadrant.

R_y = -2.8 + 6.4 - 0.7

R_y = 2.9 km

Diano4ka-milaya [45]3 years ago
4 0
Rx= 3.5 km

Ry= 2.9 km
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Oksana_A [137]
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6east-3west+1east+6east-7west

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Any questions please ask.
5 0
3 years ago
The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbi
SOVA2 [1]

Answer:

5.024 years

Explanation:

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r1 = 150 million km

r2 = 440 million km

let the period of asteroid orbit is T2.

Use Kepler's third law

T² ∝ r³

So,

\left ( \frac{T_{2}}{T_{1}} \right )^2=\left ( \frac{r_{2}}{r_{1}} \right )^3

\left ( \frac{T_{2}}{1} \right )^2=\left ( \frac{440}{150} \right )^3

T2 = 5.024 years

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4 0
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Which statement is not correct about force.
BlackZzzverrR [31]

Answer:

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4 0
2 years ago
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the
algol13

Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa

<u>Explanation:</u>

Given -

Stress Direction, A = [1 0 0 ]

Slip plane = [ 1 1 1]

Normal to slip plane, B = [ 1 1 1 ]

Critical stress, Sc = 2.92 MPa

Let the direction of slip on = [ 1 1 0 ]

Let Ф be the angle between A and B

cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3

cos Ф = 1/√3

σ = Sc / cosФ cosλ

For slip along [ 1 1 0 ]

cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1

cos λ = 1/√2

Therefore,

σ = 2.92 / 1/√3 1/√2

σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa

Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa

 

4 0
3 years ago
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8 0
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