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gulaghasi [49]
3 years ago
9

Which is a valid velocity reading for an object?

Physics
2 answers:
pashok25 [27]3 years ago
5 0

Answer:

45 m / s North is a valid vector reading for an object.

Explanation:

Then velocity will be defined by x km / hr North. And, magnitude of velocity defines the speed of the body. Although this tells the speed, but there is no description for the direction, so it's not a vector reading

zloy xaker [14]3 years ago
3 0

Answer:

Its 45 m/s north

Explanation:

cause its valid

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Aspirin is a compound composed of carbon, hydrogen, and oxygen atoms
vekshin1

Answer:

A homogeneous Mixture

Explanation:

The acid that contains the acetylsalicylic acid is a <u>mixture,</u> but it isnt a compound. though aspirin is. (hopefully this helps? qwq)

8 0
2 years ago
The velocity of sound apparatus is used in an investigation to determine the frequency of an unknown tuning fork. The temperatur
Hitman42 [59]

Answer:

Explanation:

The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves: vw = fλ, where vw is the speed of sound, f is its frequency, and λ is its wavelength.

6 0
3 years ago
An astronaut is in space with a baseball and a bowling ball. The astronaut pushes both objects in the same direction. If both ba
Crazy boy [7]

Answer:

As a mass greater than that of baseball, at the moment of the bowling wave the moment of the baseball ball is also greater

Explanation:

This problem is an application of momentum and momentum. When the astronaut pushed balls, he needed more force to move the ball of greater mass (bowling). The expression for soul is

      p = m v

Besibol Blade

      p1 = m1 v

Bowling ball

      p2 = m2 v

As a mass greater than that of baseball, at the moment of the bowling wave the moment of the baseball ball is also greater

      p2 >> p1

3 0
3 years ago
Explain how the basic unit are combined to give the derived units of force, velocity, pressure and work
LuckyWell [14K]

Velocity:

Velocity is change in displacement with respect to time:

\frac{\Delta x}{\Delta t}

Analysing the units, meters (displacement) and seconds (time) are basic units:

\frac{m}{s}

Therefore the unit of velocity is m/s

Force:

Newton's second law of motion:

F = ma

Kilogram (mass) is a basic unit, and accelerations unit can be found using the equation:

a=\frac{\Delta v}{\Delta t}

Analysing the units:

\frac{\frac{m}{s}}{s}=\frac{m}{s^2}

Therefore, the unit of force is:

kg\frac{m}{s^2}

Pressure:

Pressure is given by the equation:

P=\frac{F}{S} where S is area of effect, F is force

Area for a basic rectangle (geometric shape is arbitrary for dimensional analysis) is found by multiplying two lengths:

[l^2]=m^2, the unit of area

Dividing the aforementioned unit of force by the unit of area:

\frac{kg\frac{m}{s^2}}{m^2}=\frac{kg}{ms^2}, the unit of pressure

Work:

Work is given by the equation:

W=\vec{F}\cdot \vec{x}, (dot product may be assumed as normal multiplication for the purposes of unit analysis)

Knowing displacement's (x) unit is m:

[W]=\frac{kgm}{s^2}m=\frac{kgm^2}{s^2}, the unit of work.

3 0
3 years ago
A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

#SPJ4

7 0
2 years ago
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