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Alenkinab [10]
2 years ago
9

If the sum of all the forces acting on a moving object is zero, the object will:

Physics
1 answer:
natita [175]2 years ago
4 0

Answer:

continue moving with constant velocity

Explanation:

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Which term describes the light-sensitive structures found on the retina?
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What is the angle θ between vectors A⃗ and B⃗ if A⃗ =4ı^−4ȷ^ and B⃗ =−5ı^+7ȷ^?
ololo11 [35]

The characteristics of the scalar product allows to find the angle between the two vectors is:

  • The angle θ = 170º

The scalar product is the product between two vectors whose result is a scalar.

            A . B = |A|  |B| cos θ

Where A and B are the vectors, |A| and |B| are the modules of the vectors and θ at the angle between them.

The vector is given in Cartesian coordinates and the unit vectors in these coordinates are perpendicular.

            i.i = j.j = 1

            i.j = 0

            A . B = (4 i - 4j). * -5 i + 7j)

            A . B = - 4 5 - 4 7

            A. B = -48

We look for the modulus of each vector.

           |A| = \sqrt{x^2 +y^2 }

           |A| = \sqrt{4^2 + 4^2}  

           |A| = 4 √2

          |B| = \sqrt{5^2 +7^2}

          |B| = 8.60

We substitute.

            -48 = 4√2  8.60  cos θ

            -48 = 48.66 cos θ

            θ = cos⁻¹   \frac{-48}{48.664}  

            θ = 170º

In conclusion using the dot product we can find the angle between the two vectors is:

  • the angle θ = 170º

Learn more about the scalar product here:  brainly.com/question/1550649

8 0
3 years ago
Helpon
statuscvo [17]

Answer:

huh ?ion understand

Explanation:

have a bad day <3

7 0
3 years ago
For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with d
Sedaia [141]

Answer:

The maximum mass the bar can support without yielding = 32408.26 kg

Explanation:

Yield stress of the material (\sigma) = 200 M Pa

Diameter of the bar = 4.5 cm = 45 mm

We know that yield stress of the bar is given by the formula

                Yield Stress = \frac{Maximum load}{Area of the bar}

⇒                                \sigma = \frac{P_{max} }{A}  ---------------- (1)

⇒ Area of the bar (A) = \frac{\pi}{4} ×D^{2}

⇒                            A  = \frac{\pi}{4} × 45^{2}

⇒                            A = 1589.625 mm^{2}

Put all the values in equation (1) we get

⇒ P_{max} = 200 × 1589.625

⇒ P_{max} = 317925 N

In this bar the P_{max} is equal to the weight of the bar.

⇒ P_{max} = M_{max} × g

Where M_{max} is the maximum mass the bar can support.

⇒ M_{max} = \frac{P_{max} }{g}

Put all the values in the above formula we get

⇒ M_{max} = \frac{317925}{9.81}

⇒ M_{max} = 32408.26 Kg

There fore the maximum mass the bar can support without yielding = 32408.26 kg

3 0
4 years ago
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