All categories

3 years ago

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:

You might be interested in

anastassius [24]

Answer:

the answer is c

Explanation:

7 0

3 years ago

An 877 kg drag race car accelerates from rest to 116 km/h in 0.951 s. What change in momentum does the force produce? Answer in

Eduardwww [97]

Answer:

The change in momentum is 28265.71 kg-m/s.

Explanation:

Given that,

Mass of a car, m = 877 kg

Initial velocity of the car, u = 0 (at rest)

Final velocity of the car, v = 116 km/h = 32.23 m/s

Time, t = 0.951 s

We need to find the change in momentum produced by the force. It can be calculated as the difference of final momentum and the initial momentum.

$\Delta P=m(v-u)\\\\=877\times (32.23-0)\\\\=28265.71\ kg-m/s$

So, the change in momentum is 28265.71 kg-m/s.

7 0

2 years ago

What is massand its si unit

natulia [17]

A large body of matter with no definite shape.
The SI unit is-kilogram(kg).

6 0

3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago

The data table shows how the amplitude of a mechanical wave varies with the energy it carries. Analyze the data to identify the

Maslowich

The energy of a wave is directly proportional to the square of the amplitude of the wave.

<h3>What is the relationship between energy and amplitude?</h3>

There is direct relationship between energy of the wave and the amplitude of the wave. The energy transported by a wave is directly proportional to the square of the amplitude of the wave. This means if energy is increase the amplitude of wave becomes double and vice versa.

Energy = (amplitude)2

So we can conclude that the energy of a wave is directly proportional to the square of the amplitude of the wave.

Learn more about energy here: brainly.com/question/13881533

#SPJ1

3 0

2 years ago

Read 2 more answers