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3 years ago

What are the first two steps for finding the magnitude of the resultant vector?

Physics

1 answer:

Marina86 [1]3 years ago

8 0

Answer:

In the analytical method,

Resolve the vectors into the perpendicular components of the Cartesian coordinates.
Calculate the magnitude of the resultant vector using the Pythagoras theorem.

Explanation:

There are two methods to find the magnitude of the resultant vector.
One is the geometrical method and the other one is the analytical method.
In the geometrical method, all the vectors are connected the head to tail with the appropriate magnitude and the resultant vector is obtained by joining the initial point and the final point by a vector in the reverse direction. The magnitude of the resultant vector is given by the length of the line.
In the analytical method, all the vectors are resolved into the perpendicular components.
Using Pythagoras theorem, the magnitude of the resultant vector can be obtained
If A and B are the two vectors forming an angle ∅ between them, then the magnitude of the resultant vector is given by the formula

$R=\sqrt{A^{2}+B^{2}+2ABcos\phi}$

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A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule

saw5 [17]

Explanation:

Mass of the astronaut, m₁ = 170 kg

Speed of astronaut, v₁ = 2.25 m/s

mass of space capsule, m₂ = 2600 kg

Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :

initial momentum = final momentum

Since, initial momentum is zero. So,

$m_1v_1+m_2v_2=0$

$170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0$

$v_2=-0.17\ m/s$

So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.

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3 years ago

Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw

Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

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3 years ago

Which is more dense, cold freshwater or warm seawater?

svp [43]

Cold freshwater is denser than warm seawater, because of the salinity and temperature variations. Cold water would have less salt since the solubility of the salt is lower as compared to warm water. Hope this answers the question. Have a nice day.

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3 years ago

Read 2 more answers

The speed of light in a solid is 1.24 x 108 m/s. Calculate the index of refraction

Dahasolnce [82]

Answer:

125

Explanation:

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2 years ago

A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a

EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is: