1answer.
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
3241004551 [841]
3 years ago
11

The distance vs. time graph of a car moving at constant speed should be a straight line. Why do the data points in the graph plo

tted from observing the motion not fall right on the line? A graph with 7 points and a line segment representing a straight line. Horizontal axis is labeled Time, in seconds, and extends from 0 to 7. Vertical axis is labeled Distance, in meters, and extends from 0 to 50. Approximate coordinates of plotted points are begin ordered pair 1 comma 8 end ordered pair comma begin ordered pair 2 comma 13 end ordered pair comma begin ordered pair 3 comma 21 end ordered pair comma begin ordered pair 4 comma 28 end ordered pair comma begin ordered pair 5 comma 35 end ordered pair comma begin ordered pair 6 comma 39 end ordered pair comma begin ordered pair 7 comma 49 end ordered pair. Line segment has positive slope, and it begins at the point begin ordered pair 0 comma 0 end ordered pair, and it ends at the point begin ordered pair 7 comma 49 end ordered pair. The measurements are inexact. Data points can never look like they fall right on a straight line. The distance is not exactly proportional to time even if the speed is exactly constant. The distance and time are proportional only if the correct units of meters and seconds are used.
Physics
1 answer:
iragen [17]3 years ago
6 0

Answer:

The measurements are inexact

Explanation:

You might be interested in
Answer the following questions
zaharov [31]

Answer:

9 - 10N to the left

10 - There is no change on the object

Explanation:

Can I have brainliest answer pls?

7 0
2 years ago
Read 2 more answers
A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points
mr Goodwill [35]

Answer:

Approximately 11.0\; \rm m \cdot s^{-1}. (Assuming that g = 9.81 \; \rm N \cdot kg^{-1}, and that the tabletop is level.)

Explanation:

Weight of the book:

W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N.

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, F(\text{normal force}) \approx 10.202\; \rm N.

As a side note, the F_N and W on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction F(\text{kinetic friction}) on it will be equal to

  • \mu_{\rm k}, the coefficient of kinetic friction, times
  • F(\text{normal force}), the normal force that's acting on it.

That is:

\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}.

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that 15.0\; \rm N applied force. The net force on the book shall be:

\begin{aligned}& F(\text{net force})  \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}.

Apply Newton's Second Law to find the acceleration of this book:

\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}.

6 0
3 years ago
A dog running at 10 m/s is 30m behind a rabbit moving at 5 m/s. when will the dog catch up with the rabbit assuming both their v
dedylja [7]

The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.

<h3>What time will the dog catch the rabbit?</h3>

The time that the dog will catch up with the rabbit is given as follows:

Let the distance covered by the rabbit be x.

Distance covered by dog = x + 30

  • Time taken = distance/speed

The time taken will be the same T

  1. Time taken by dog, T = (x + 30)/10
  2. Time taken by rabbit, T = x/5

Equating both times.

(x + 30)/10 = x/5

x = 30 m

Solving for T in equation (ii);

T = 30/5 = 6 minutes

In conclusion, time is obtained as a ratio of distance and speed.


Learn more about time and speed at: brainly.com/question/26046491

#SPJ1

7 0
2 years ago
When a wave is bent by traveling from one medium to another
Natalka [10]

Answer:

B. Refraction

Explanation:

Jope this helps

6 0
3 years ago
Read 2 more answers
The force of attraction between a ball is F=.........×10^-¹¹
DIA [1.3K]

Answer:

4.45×10¯¹¹ N

Explanation:

From the question given above, the following data were obtained:

Mass of ball (M₁) = 4 Kg

Mass of bowling pin (M₂) = 1.5 Kg

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Distance apart (r) = 3 m

Force of attraction (F) =?

The force of attraction between the ball and the bowling pin can be obtained as follow:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 4 × 1.5 / 3²

F = 4.002×10¯¹⁰ / 9

F = 4.45×10¯¹¹ N

Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N

8 0
3 years ago
Other questions:
  • What is a continuous range of a single feature such as a wave lengt
    8·1 answer
  • A person in a kayak starts paddling, and it accelerates from 0 to 0.65 m/s in a distance of 0.40 m. If the combined mass of the
    15·1 answer
  • What is a natural form of pollution?
    12·1 answer
  • Two trains leave the station at the same time, one heading east and the other west. the eastbound train travels at 85 miles per
    14·1 answer
  • 2. Consider a conical pendulum with a bob of mass m = 77.0 kg on a string of length L = 10.0 m that makes an angle of ???? 3.00°
    8·1 answer
  • A 10 kg block is pushed with a constant horizontal force of 20 n against a constant horizontal frictional force of 10 n. What is
    9·1 answer
  • Ricardo, mass 85 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 20 kg canoe. When the canoe is at rest
    11·1 answer
  • Which type of motion occur in a lift in one word<br>Pls answer​
    14·1 answer
  • 2 part question
    5·1 answer
  • a 4,000 kilogram rocket has accelerates at a rate of 35 m/s2. How much force is required to do this?​
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!