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lisov135 [29]
3 years ago
11

A student collected data about the distance a car travels over time.

Physics
2 answers:
Molodets [167]3 years ago
5 0

Answer:

Therefore the correct answer is B.  Scatterplot

Explanation:

This is an exercise in kinematics, in a measurement of distance against time, the process is described by the expression

          x = vo t + ½ a t2

where vo is the initial velocity. At the acceleration of the body, x the displacement and t the time.

Depending on the type of movement, a can be zero or different from zero.

In general, all these data are graphed in a DISPERSION graph where the distance is on the abscissa axis and the ordinate axis is time.

Therefore the correct answer is B

katovenus [111]3 years ago
4 0

Answer:

It's B. Scatterplot I just took the test

Explanation:

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a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

4 0
3 years ago
In January 2006, astronomers reported the discovery of a planet comparable in size to the earth orbiting another star and having
Orlov [11]

Answer:

R = 5.28  103 km

Explanation:

The definition of density is

              ρ = m / V

              V = m /ρ

Where m is the mass and V the volume of the body

The volume of a sphere is

            V = 4/3 π r³

Let's replace

             4/3 π r³ = m / ρ

             R =∛ ¾ m / ρ π

The mass of the planet is

              M = 5.5 Me

              R = ∛ ¾ 5.5 Me /ρ π

Let's reduce the density to SI units

             ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³

             ρ = 1.76 10³ kg / m³

Let's calculate

               R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)

               R = ∛ 0.14723 10²¹

               R = 0.528 10⁷ m

               R = 0.528 104 km

               R = 5.28  103 km

8 0
3 years ago
You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock
kondor19780726 [428]

Answer:

0.4778 m/s

Explanation:

To solve this question, we will make use of law of conservation of momentum.

We are given that the rock's velocity is 12 m/s at 35°. Thus, the horizontal component of this velocity is;

V_x = (12 m/s)(cos(35°)) = 9.83 m/s.

Thus, the horizontal component of the rock's momentum is;

(3.5 kg)(9.83 m/s) = 34.405 kg·m/s.

Since the person is not pushed up off the ice or down into it, his momentum will have no vertical component and so his momentum will have the same magnitude as the horizontal component of the rock's momentum.

Thus, to get the person's speed, we know that; momentum = mass x velocity

Mass of person = 72 kg and we have momentum as 34.405 kg·m/s

Thus;

34.405 = 72 x velocity

Velocity = 34.405/72

Velocity = 0.4778 m/s

6 0
3 years ago
A measure of how far an object has moved from a starting point
Anvisha [2.4K]
Volume??? velocity??????
5 0
3 years ago
A car drives at steady speed around a perfectly circular track.
gayaneshka [121]

Answer:

e. Both the acceleration and net force on the car point inward.

Explanation:

If no net force acts on the car, the car must drive in a straight line, at constant speed.

As the acceleration is defined as the rate of change of the velocity vector, this means that it can produce either a change in the magnitude of the velocity (the speed) or in the direction.

In order to the car can follow a circular trajectory, it must be subjected to an acceleration, that must go inward, trying to take the car towards the center of the circle.

The net force that causes this acceleration, aims inward, and is called the centripetal force.

It is not a different type of force, it can be a friction force, a tension force, a normal force, etc., as needed.

6 0
3 years ago
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