answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Step2247 [10]
2 years ago
6

Rosa studies the position-time graph of two race cars. A graph titled Position versus Time shows time in hours on the x axis, nu

mbered 0 to 5, position in kilometers on the y axis, numbered 0 to 50. The graph has two straight lines, one from point (0, 20) up to (2.8, 50), the second line starts at the origin and goes to point (5, 20). Rosa concludes that truck X has a greater velocity than truck Y, and that truck Y had a head start. She also concluded that the graph shows both trucks moving at a constant velocity and that neither truck changes direction. Which statement best describes Rosa’s error?
A Both trucks change direction after they start moving.
B Truck X had a head start, not truck Y.
C Truck Y has a greater velocity than truck X.
D The trucks are not traveling at constant velocities.
Physics
2 answers:
Harlamova29_29 [7]2 years ago
5 0

Answer:

B. Truck X had a head start, not truck Y.

Explanation:

Let's see what data can we obtain from this information.

Truck X's position changed from point (0,20) to point (2.8,50). That means that it started on the 20th kilometer and in the next 2.8 hours reached 50 km. It's velocity, then, is v1 = (s2 - s1) / t

v1 = (50 - 20) / 2.8

v1 = 10.7 km/h

Since it started from 20th km that means it had head start and since its line on the graph is straight, that means that its velocity was constant and that it didn't change direction.

Truck Y's position changed from the origin (0,0) to the point (5,20). That means it crossed 20 km in 5 hours, so its velocity is v2 = 20 / 5

v2 = 4 km/h

Again, since its line on the graph is straight, its velocity is constant and it did not change direction.

Now, knowing this, we can see that Rosa's mistake was to claim that truck Y had head start.

fgiga [73]2 years ago
4 0

Answer:

Answer is C

Explanation:

Just took the test and got it right

You might be interested in
A body is projected upward at an angle of 30 degree to the horizontal at an initial speed of 200ms-.In how many seconds will it
Crazy boy [7]

Answer:

20.41 s

3534.80 m

Explanation:

<em><u>In how many seconds will it reach the ground?</u></em>

We are given the initial velocity of the body, which is 200 m/s at a 30° angle.

We know the acceleration in the vertical direction is -9.8 m/s², assuming that the upwards/right direction is positive and the downwards/left direction is negative.

Since we are using acceleration in the y-direction, let's use the vertical component of the initial velocity.

  • 200 · sin(30) m/s

Let's use the fact that at the top of its trajectory, the body will have a final velocity of 0 m/s.

Now we have one missing variable that we are trying to solve for: time t.

Find the constant acceleration equation that contains v₀, v, a, and t.

  • v = v₀ + at

Substitute known values into the equation.

  • 0 = 200 · sin(30) + (-9.8)t
  • -200 · sin(30) = -9.8t
  • t = 10.20408163

Recall that this is only half of the body's trajectory, so we need to double the time value we found to find the total time the body is in the air.

  • 2t = 20.40816327

The body will reach the ground in 20.41 seconds.

<em><u>How far from the point of projection would it strike? </u></em>

We want to find the displacement in the x-direction for the body.

Let's find the constant acceleration equation that contains time t, that we just found, and displacement (Δx).

  • Δx = v₀t + 1/2at²

Substitute known values into the equation. Remember that we want to use the horizontal component of the initial velocity and that the acceleration in the x-direction is 0 m/s².

  • Δx = (200 · cos(30) · 20.40816327) + 1/2(0)(20.40816327)²
  • Δx = 3534.797567

The body will strike 3534.80 m from the point of projection.

4 0
1 year ago
On a ring road, 12 trams are spaced at regular intervals and travel at a constant speed. How many trams need to be added to the
Sphinxa [80]

3 trams must be added

Explanation:

In this problem, there are 12 trams along the ring road, spaced at regular intervals.

Calling L the length of the ring road, this means that the space between two consecutive trams is

d=\frac{L}{12} (1)

In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become

d'=(1-\frac{1}{5})d=\frac{4}{5}d

And the number of trams will become

12+n

So eq.(1) will become

\frac{4}{5}d=\frac{L}{n+12} (2)

And substituting eq.(1) into eq.(2), we find:

\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3

Learn more about distance and speed:

brainly.com/question/8893949

#LearnwithBrainly

4 0
2 years ago
A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At
GuDViN [60]

Answer:

E/4

Explanation:

The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

Where;

E is the electric field

σ is the surface charge density

ε₀ is the electric constant.

Formula to calculate σ is;

σ = Q/A

Where;

Q is the total charge of the sheet

A is the sheet's area.

We are told the elastic sheet is a square with a side length as d, thus ;

A = d²

So;

σ = Q/d²

Putting Q/d² for σ in the electric field equation to obtain;

E = Q/(2ε₀d²)

Now, we can see that E is inversely proportional to the square of d i.e.

E ∝ 1/d²

The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.

From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;

E_new = E/4

3 0
2 years ago
Which of these is the most effective way for Leanna to cool down after an intense bike ride
Sonja [21]
I am pretty sure the answer would be too stretch
6 0
2 years ago
3. A 4.1 x 10-15 C charge is able to pick up a bit of paper when it is initially 1.0 cm above the paper. Assume an induced charg
Anni [7]

Answer:

\mathbf{1.51\times10^{-15}N}

Explanation:

The computation of the weight of the paper in newtons is shown below:

On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.

Therefore the paper charge is

q_{paper}=-4.1\times10^{-15}C

Now the distance from the charge is

r=1cm=0.01m

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.

mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}

\Rightarrow W=mg

=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}

=\mathbf{1.51\times10^{-15}N}

6 0
1 year ago
Other questions:
  • Can someone with an IQ score of 120 be gifted? Based on psychological thought, explain why or why not and give an example. ILL M
    16·2 answers
  • A circular loop of wire is rotated at constant angular speed about an axis whose direction can be varied. In a region where a un
    7·1 answer
  • A 0.468 g sample of pentane, C 5H 12, was burned in a bomb calorimeter. The temperature of the calorimeter and 1.00 kg of water
    12·1 answer
  • Sketch the circuit labeling the meter and bulb as two separate resistors connected in parallel to the voltage source. Then show
    6·1 answer
  • || Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and r
    14·2 answers
  • Two parallel plates are a distance apart with a potential difference between them. a point charge moves from the negatively char
    7·1 answer
  • A team of engineering students is testing their newly designed raft in the pool where the diving team practices.
    13·1 answer
  • A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled
    7·1 answer
  • A cross country skier moves from location A to location B to location C to location D. Each leg of the back and forth motion tak
    7·1 answer
  • A 3400 kg jet is flying at a constant speed of 170 m/s as it makes a vertical loop. At the top of the loop the pilot feels three
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!