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MA_775_DIABLO [31]
3 years ago
6

Which of the following are vector quantities? Select all that apply.

Physics
1 answer:
valkas [14]3 years ago
8 0

Answer:

7 miles southwest

9 steps to the left

75 mph north

Explanation:

A vector quantity is any physical quantity that specifies the magnitude of the quantity and its direction.

So, the choices that gives a magnitude and direction are vectors.

The other type of physical quantities are the scalar which only specifies the magnitude but no direction;

9 steps to the left and  7 miles southwest are called displacement

75 mph north is velocity

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Find the position of the center of mass of the system of the sun and Jupiter? (Since Jupiter is more massive than the rest of th
8090 [49]

Answer:

r_{cm} = 0.074 m from the position of the center of the Sun

Explanation:

As we know that mass of Sun and Jupiter is given as

M_s = 1.98 \times 10^{30} kg

M_j = 1.89 \times 10^{27} kg

distance between Sun and Jupiter is given as

r = 7.78 \times 10^{11} m

now let the position of Sun is origin and position of Jupiter is given at the position same as the distance between them

so we will have

r_{cm} = \frac{M_s r_1 + M_j r_2}{M_s + M_j}

r_{cm} = \frac{1.98 \times 10^{30} (0) + (1.89 \times 10^{27})(7.78 \times 10^{11})}{1.98 \times 10^{30} + 1.89 \times 10^{27}}

r_{cm} = 0.074 m from the position of the center of the Sun

3 0
3 years ago
In a solution, the solvent is present in the larger amount. <br> A True <br> B False
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8 0
3 years ago
An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50 newtons. Calcuate the acceleration
rjkz [21]

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this answer you question

6 0
3 years ago
A(n) 1946 kg car travels at a speed of 10 m/s . What is its kinetic energy ? Answer in units of J.
erica [24]

Answer:

KE=97300J

Explanation:

KE=1/2mv^2

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KE=97300J

4 0
3 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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