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love history [14]
3 years ago
6

Two vectors A and B are added together to form a vector C. The relationship between the magnitudes of the vectors is given by A

+ B = C. Which one of the following statements concerning these vectors is true?
Physics
1 answer:
il63 [147K]3 years ago
4 0

Answer:

The question is incomplete, below is the complete question

"Two vectors A and B are added together to form a vector C. The relationship between the magnitudes of the vectors is given by A + B = C. Which one of the following statements concerning these vectors is true?

a. A and B must be displacements.

b. A and B must have equal lengths.

c.A and B must point in opposite direction.

d. A and B must point in the same direction.

e.A and B must be at right angles to each other."

answer:

d. A and B must point in the same direction.

Explanation:

a.false:From vector analysis, all forms of vector with the same unit can be added, and we add vector component by component. hence this defile option (a).

b. false: in addition of vectors, length is not a criteria to consider before carrying out the operation, hence vectors of different lengths can be  added

c. false: this is against the claim as vectors in opposite direction can give rise to a new vector of negative value.

d. true: this is true as vector in the same direction add up to give rise to another vector.

e.false: this is not a true assumption

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C. Forces cannot be quantified
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2 years ago
A block with a mass of 1kg moving at a velocity of 3m/s collides and sticks to a block of mass of 4kg initially at rest. What is
Vsevolod [243]

Answer:using Newton third law

Let initial velocity of block be u1=3m/s

Mass of moving block m1 =1kg

Final velocity of block =V

Mass of stationary block m2= 4kg

Since they stick together, their final velocity will be the same.

m1u1 + m2u2=(m1+m2)v

(1*3)+(0*4)=(1+4)v

3=5v

Divide both sides by 5

V=0.6

Final velocity is 0.6m/s

Explanation:

8 0
3 years ago
The slotted arm OA rotates about a fixed axis through O. At the instant under consideration, θ = 36°, θ˙ = 36 deg/s, and θ¨ = 26
xz_007 [3.2K]

Answer:

   F = 0.0545 N

Explanation:

Let's use Newton's second law for rotational movement

        τ = I α

The moment of inertia for a bar supported by some ends is

       I = 1/3 m L²

Torque is

            τ = F L

Let's replace

         F L = 1/3 m L² α

         F = 1/3 m L α

Let's reduce angular acceleration to SI units

       Alf = 26º / s² (π rad / 180º) = 0.454 rad / s²

Let's calculate

          F = 1/3  0.5  0.72  0.454

          F = 0.0545 N

4 0
3 years ago
What words from the vocabulary that is on the bottom left corner go in number 4-8?
Helen [10]

Answer:

4) Inner Core

5) Outer Core

6)Crust

7) Convection

8)   Volcanoes    

Explanation:

5 0
2 years ago
Note: The rope is 20 m long. Answer like this: (1.<br> 2._____ etc)
qaws [65]

Answer:

1. <u>Potential energy</u>, 2. <u>Potential and kinetic energy</u>, 3. <u>Potential and kinetic energy</u>, 4. <u>Potential and kinetic energy</u>, 5. <u>Potential energy</u>

Explanation:

We note that the total mechanical energy (M.E.) of the body is given as follows;

M.E. = K.E. + P.E. = Constant

Where;

K.E. = The kinetic energy of the body = (1/2)·m·v²

P.E. = The potential energy of the body = m·g·h

m = The mass of the person

v = The velocity with which the person is in motion

g = The acceleration due to gravity ≈ 9.81 m/s²

h = The height of the person above the ground

The length of the rope = 20 m

The initial height at location 1, h₁ = 40.0 m

At location 1, the velocity, v₁ = 0.00 m/s

The mechanical energy, M.E. = K.E.₁ + P.E.₁

∴  K.E.₁ = 0 and P.E.₁ = m ×9.81×40

M.E. = (1/2) ×m ×0² + m ×9.81×40

∴ M.E. = 0 + P.E.₁ the type of energy present at location 1 is only potential energy

At location 2, the velocity, v₂ = 10.0 m/s

The mechanical energy, M.E. = K.E.₂ + P.E.₂ = (1/2) ×m ×10² + m ×9.81×40

∴  K.E.₂ = 50·m and P.E.₂ = m ×9.81×35 = 343.35·m

M.E. = 50·m + 343.35·m the type of energy at location 2 is both kinetic energy, K.E. and potential energy, P.E.

At location 3, the velocity, v₃ = 20.0 m/s

The mechanical energy, M.E. = K.E.₃ + P.E.₃ = (1/2) ×m ×20² + m ×9.81×20

∴  K.E.₃ = 200·m and P.E.₃ = m ×9.81×20 = 196.2·m

M.E. = 200·m + 196.2·m the type of energy at location 3 is both kinetic energy, K.E. and potential energy, P.E.

At location 4, the velocity, v₄² = 350.0 m²/s², h₄ = 15.0 m

The mechanical energy, M.E. = K.E.₄ + P.E.₄ = (1/2) × m ×350 + m ×9.81×15

∴  K.E.₄ = 175·m and P.E.₄ = m×9.81×15 = 147.15·m

M.E. = 175·m + 147.15·m the type of energy at location 4 is both kinetic energy, K.E. and potential energy, P.E.

At location 5, the velocity, v₅ = 0 m/s, h₅ = 10.0 m

The mechanical energy, M.E. = K.E.₅ + P.E.₅ = (1/2) × m × 0 + m ×9.81×10

∴  K.E.₅ = 0·m and P.E.₅ = m×98.1 = 98.1·m

M.E. = 0·m + 98.1·m the type of energy at location 5 is only potential energy, P.E.

Therefore, we have;

\left|\begin{array}{ccc}Location&&Type(s) \ of \ Energy \ Presents\\1&&Potential \ Energy\\2&&Potential  \ and \ Kinetic \ Energy\\3&&Potential  \ and \ Kinetic \ Energy\\4&&Potential  \ and \ Kinetic \ Energy\\5&&Potential  \  Energy\end{array} \right |

5 0
2 years ago
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