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3 years ago

The magnitude of a vector can never be less than the magnitude of one of its components. Group of answer choices True False

Physics

1 answer:

pogonyaev3 years ago

6 0

Answer:

True

Explanation:

As the formula for magnitude of a vector v and their components $v_1, v_2, v_3, v_4,..., v_n$ is

$v^2 = v_1^2 + v_2^2 + ... +v_n^2$

Since $v_1^2, v_2^2, v_3^2, v_4^2,..., v_n^2 \geq 0$ , this means the sum of them, $v^2$ , is always greater or equal to $v_1^2, v_2^2, v_3^2, v_4^2,..., v_n^2$

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What is my displacement if i walked 200 meters west then turned around and walked 150 meters east

Alinara [238K]

You're walking in one direction, and then the exact opposite of that direction, so you simply have to subtract the two distances.
200-150=50
You're 50 meters west of where you originally started.
You're west because 200 meters west is greater than 150 meters east. If the distance walked east was greater than the distance walked west, you would've been east of your starting position.

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2 years ago

The density of a substance equals its mass divided by its volume. Talia listed the density of some common materials at 20 °C.

Zolol [24]

Answer:

mercury

Explanation:

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2 years ago

Read 2 more answers

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coeffi

Flura [38]

The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and $\mu$ be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

v = 0, s = 84 m, $\mu$ = 0.36

According to Newton's law of second motion the expression for acceleration is as follows.

F = ma

$-\mu N$ = ma

$-\mu mg$ = ma

a = $-\mu g$

Also,

$v^{2} = u^{2} + 2as$

$(0)^{2} = u^{2} + 2(-\mu g)s$

$u^{2} = 2(\mu g)s$

= $\sqrt{2(0.36)(9.81 m/s^{2})(84 m)}$

= 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

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2 years ago

The perception of high or low sounds is called ____________. 2. The _______________ is a measurement unit for intensity of sound

matrenka [14]

1. pitch

2. decibel

3. sorry, i dont know

4. c

5. acoustics

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2 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f

Alborosie

Answer:

3.25 m

Explanation:

t = Time taken = 0.166 seconds

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s²

s = 1 because meter stick is 1 meter in length

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1-\frac{1}{2}\times 9.81\times 0.166^2}{0.166}\\\Rightarrow u=5.21\ m/s$

Here, the initial velocity of point B is calculated from the time which is given. This velocity will be the final velocity of the acorn which falling from point A.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.21^2-0^2}{2\times 9.81}\\\Rightarrow s=1.38\ m$

The distance of the acorn from the ground is 1.87+1.38 = 3.25 m

4 0

2 years ago