All categories

3 years ago

The and _ are used to calculate magnitude and direction of a resultant vector.

1 answer:

3 0

Here are the correct answers that would complete the given statement above. The vector quantity and the vector arrow are used to calculate magnitude and direction of a resultant vector. Vector quantity has both magnitude and direction, whereas vector arrow represents<span> the magnitude of a quantity and the direction represents the direction of that quantity. </span>Hope this is the answer that you are looking for.

You might be interested in

A vector is 14.4 m long and

MaRussiya [10]

Answer:

Explanation:

The x-component is found in the magnitude of the vector times the cosine of the angle.

$A_x=14.4cos133$ and, to 3 sig dig,

$A_x=-9.82m$

3 0

3 years ago

At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counsel

sleet_krkn [62]

Answer:

47 m

Explanation:

Data obtained from the question include the following:

Length of dry leg 1 (L1) = 40 m

Length of dry leg 2 (L2) = 25 m

Length of swimming course (L) =..?

The length of the swimming course can be obtained by using pythagoras theory as shown below:

L² = L1² + L2²

L² = 40² + 25²

L² = 1600 + 625

L² = 2225

Take the square root of both side.

L = √2225

L = 47.1 ≈ 47 m

Therefore, the length of the swimming course is approximately 47 m.

7 0

3 years ago

Is the energy of a wave is calculated by squaring the frequency of the wave?

Misha Larkins [42]

Nope.
Energy is directly proportional to frequency. and when you calculate energy, you multiply frequency with a constant number called "Planck's Constant"

E = hf

Hope this helps!

5 0

3 years ago

If something is a good conductor, what type of insulator is it?

nlexa [21]

4. a poor insulator

If rest other things are kept constant or unchanged then a good conductor can be termed as a poor insulator.

8 0

1 year ago

Read 2 more answers

A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of

posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

By substituting the values, we get

$\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}$

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

6 0

3 years ago

The ______ and _______ are used to calculate magnitude and direction of a resultant vector.

The and _ are used to calculate magnitude and direction of a resultant vector.