Physics, 07.03.2021 15:50 Nbeans4962

Find the magnitude of the sum of these two vectors:
5.00 mA,
30.0°
B4
6.00 m
magnitude (m)

Find the magnitude of the sum

of these two vectors:5.00 mA,30.0°B46.00 mmagnitude (m)

Answers: 1

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Answers

Answer from: shainaanderson24

Approximately $3.01\; \rm m$ .

Explanation:

Decompose each vector into the sum of two vectors: a horizontal one (parallel to the arrow that points to the right) and a vertical one (parallel the arrow that points upwards.)

Vector $\sf A$ is horizontal and is at an angle of $0^\circ$ with the horizon.

Horizontal component of vector $\sf A$ : to the right, with a length of $5.00\; \rm m \cdot \cos\left(0^\circ \right) = 5.00\; \rm m$ .Vertical component of vector $\sf A$ : $5.00\; \rm m \cdot \sin\left(0^\circ \right) = 0\; \rm m$ .

Vector $\sf B$ is at an angle of $30^\circ$ below the horizon.

Horizontal component of vector $\sf B$ : to the right, with a length of $\displaystyle 6.00\; \rm m \cdot \cos\left(30^\circ \right) = (6.00\; \rm m) \times \frac{\sqrt{3}}{2}\approx 5.19615\; \rm m$ .Vertical component of vector $\sf B$ : downwards, with a length of $\displaystyle 6.00\; \rm m \cdot \sin\left(30^\circ \right) = 6.00\; \rm m \times \frac{1}{2} = 3.00\; \rm m$ .

Calculate the sum of vector $\sf A$ and vector $\sf B$ .

The horizontal component of vector $\sf A$ and vector $\sf B$ are opposite to one another. Therefore, the length of the horizontal component of $\sf (A + B)$ would be the difference between the length of the horizontal components of vector $\sf A\!$ and of vector $\sf B\!$ :

$\displaystyle (6.00\; \rm m) \times \frac{\sqrt{3}}{2} - 5.00\; \rm m \approx 0.196152\; \rm m$ .

The length of the vertical component of vector $\sf A$ is $0\; \rm m$ . Therefore, the length of the vertical component of $\sf (A + B)$ would be equal to the length of the vertical component of vector $\sf B$ , $\displaystyle 6.00\; \rm m \times \frac{1}{2} = 3.00\; \rm m$ .

Therefore, the length of the horizontal and vertical component of $\sf (A + B)$ are approximately $0.196152\; \rm m$ and $3.00\; \rm m$ , respectively. The length of vector $\sf (A + B)\!$ would be approximately:

$\displaystyle \sqrt{(0.196152\; \rm m)^{2} + (3.00\; \rm m)^{2}} \approx 3.01\; \rm m$ .

Answer from: Quest

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.

Answer from: Quest

Velocity=total displacement/timefrom house to store is 500 from store to neighbor is 200 therefore total displacement=500+200=700metrestherefore,700metres/910seconds=0.8metres per second

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Find the magnitude of the sum of these two vectors:
5.00 mA,
30.0°
B4
6.00 m