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Fantom [35]
4 years ago
11

You travel in a circle, whose circumference is 8kilometers, at an average speed of kilometer/hour. If you stop at the same point

you started from, what is your average velocity
Physics
1 answer:
dimaraw [331]4 years ago
3 0

You skipped over a number in the question, and you didn't tell me what my average speed is. Lucky for you, my average speed has NO EFFECT on the answer to the question.

When you calculate velocity, you only use the straight-line distance between the start-point and the end-point. It doesn't matter what route the thing took to get there, or how much ground it actually covered.

If I travel in a circle and stop at the same point I started from, then the size of the circle doesn't matter, and neither does my speed. The distance between my start-point and my end-point is zero, and my average velocity is zero.

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You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
A 2kg ball is thrown upward with a velocity of 15 m/s what is the kinetic energy of the ball being thrown?
vivado [14]
Kinetic Energy, or K.E is equal to 1/2 times the mass of an object times the velocity squared.

Therefore K.E
=  \frac{1}{2}  \times 2 \times  {15}^{2}
which is 225

Energy is expressed in joules (J)
so the K.E of the ball is 225J

Hope this helped
7 0
4 years ago
The magnitude of vector A is 2.5 m and is directed towards east. The magnitude and direction of A /2 will be​
const2013 [10]

The magnitude and direction of half of the given vector A is 1.25 m east.

The given parameters;

  • <em>magnitude of vector A = 2.5 m east</em>

A vector can be described with both magnitude and direction. The magnitude of the vector is the scalar or numeric representation of the vector.

When the scalar representation has direction, we term the quantity a vector quantity.

The magnitude of half of the given vector is calculated as follows;

\frac{A}{2} = \frac{2.5 \ m}{2} = 1.25 \ m\ east

Thus, the magnitude and direction of half of the given vector A is 1.25 m east.

Learn more here:brainly.com/question/14586816

4 0
3 years ago
The __________ wave follows the depolarization wave across the sarcolemma
sammy [17]
Your answer is repolarization
3 0
3 years ago
How do scientist know how much the world really rotates​
kotykmax [81]
“The Smithsonian pendulum, like all pendulums, moved in accordance with Foucault’s sine law, which predicts how much a pendulum’s path will distort each day based on its latitude. Absent any exterior forces, a pendulum would swing back and forth in a single plane forever—there would be no gradual angular shift. But the Earth is rotating, so the story isn’t that simple.

Since all points on Earth’s surface rotate as a unit, it follows that those located on the wider portions of the planet—nearer to the equator—must cover more meters each second (i.e., go faster) to “keep up” with the points tracing smaller circles each day at the extreme northern and southern latitudes. Though they don’t feel it, a person standing in Quito, Ecuador, is moving with appreciably higher velocity than one in Reykjavik, Iceland.

Because each swing of a pendulum takes it from a point farther from the equator to a point nearer to the equator and vice versa, and the velocities at these points differ, the path of the pendulum is subtly distorted with every swing, gradually torqued away from its original orientation. The extent of this effect depends on where on Earth the pendulum is swinging.

At the North Pole—where small changes in latitude have big implications—the path traced by a pendulum would shift through a full 360 degrees in a mere 24 hours, explains Thompson. At the equator, meanwhile, a pendulum’s motion would not be seen to distort at all.” From the Smithsonian Magazine
4 0
3 years ago
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