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jekas [21]
3 years ago
8

How can vectors be used in real life?

Physics
1 answer:
7nadin3 [17]3 years ago
4 0
Using vectors can be troublesome for simple physics problems. however, when you move onto solving more difficult problems that have several variables, and the space you are dealing with is in three dimensions, using vectors is a necessity because it's not always possible to visualize the problem. but using vectors will give you a definitive answer.

few applications of vectors:
-predict the trajectory of airplanes and make sure that the airplane will not collide with some other airplane
-all simulations use vectors
-multivariable Calculus is pretty much based on vectors. without multivariable Calculus, it would have not been possible to compute many formulas.
-modeling the path a rocket will take, and supplying it the right amount of fuel so that it gets to the moon, or some other destination.
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A client experiences difficulty in performing the prone iso-abs exercise. Which of the following should be suggested as an regre
butalik [34]

Answer:

a. Quadruped arm and opposite leg raise

Explanation:

Quadruped arm and opposite leg lift

- Kneel on the floor, lean forward and place your hands down.

- Keep your knees in line with your hips and hands directly under your shoulders.

- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.

- Go back to the starting position.

This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.

It is also used together with other exercises for the treatment of hyperlordosis.

8 0
3 years ago
he adventurous robot M.A.N.D.I. is orbiting Saturn’s moon Dione. She wants to cause an impact with themoon to kick up some of th
GuDViN [60]

Answer:

v = 2.928 10³ m / s

Explanation:

For this exercise we use Newton's second law where the force is the gravitational pull force

         F = ma

         a = F / m

Acceleration is

        a = dv / dt

        a = dv / dr dr / dt

        a = dv / dr v

        v dv = a dr

We substitute

       v dv = a dr

       ∫ v dv = 1 / m G m M ∫ 1 / r² dr

We integrate

       ½ v² = G M (-1 / r)

We evaluate from the lower limit v = 0 for r = R m to the upper limit v = v for r = R + 2.73 10³, where R is the radius of Saturn's moon

       v² = 2G M (- 1 / R +2.73 10³+ 1 / R)

         

We calculate

       v² = 2 6,674 10⁻¹¹ 1.10 10²¹ (10⁻³ / 5.61  - 10⁻³ /(5.61 + 2.73))

       v² = 14.6828 10⁷ (0.1783 -0.1199)

       v = √8.5748 10⁶

       v = 2.928 10³ m / s

5 0
3 years ago
Please someone help me !!
sergij07 [2.7K]

Answer:

D

Explanation:

because if the solvent is more than the solvent then we can't resolve it.

so our product will be suspended

6 0
2 years ago
guys be careful noblepenguin is getting people banned he took all my brainliest he has to be stopped has a bot a be careful Kati
Julli [10]

Answer:

K thanks for letting everyone know!

Explanation:

7 0
3 years ago
Read 2 more answers
A force of 14 N acts on a 5 kg object for 3 seconds.
DiKsa [7]

Answer: a) 42Nm b) 8.4m/s

Explanation:

Impulse is defined as object change in momentum.

Since Force = mass × acceleration

F = ma

Acceleration is the rate of change in velocity.

F = m(v-u)/t

Cross multiply

Ft = m(v-u)

Since impulse = Ft

and Ft = m(v-u)... (1)

The object change in velocity (v-u) = Ft/m from eqn 1

Going to the question;

a) Impulse = Force (F) × time(t)

Given force = 14N and time = 3seconds

Impulse = 14×3

Impulse = 42Nm

b) The object change in velocity (v-u) = Ft/m where mass = 5kg

v-u = 14×3/5

Change in velocity = 42/5 = 8.4m/s

3 0
3 years ago
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