All categories

1 year ago

Which statement is true regarding a distance vs. time graph? (1 point)

2 answers:

Nadusha1986 [10]1 year ago

6 0

The statement that is true regarding a distance vs. time graph is option A: The graph should show distance on the vertical axis.

<h3>Where is the plot of distance?</h3>

How far an object has come in a certain amount of time is displayed on a distance-time graph. Time is represented on the X-axis and Distance is plotted on the Y-axis (left) (bottom).

On a distance-time graph, an object's motion is indicated by a sloping line. The slope or gradient of the line in a distance-time graph is equal to the object's speed. The object is travelling more quickly the steeper the line is (and the bigger the gradient).

Note that the distance-time graph shows the relationship between distance and time by plotting distance on the y-axis and time on the x-axis.

Learn more about distance vs. time graph from

brainly.com/question/16825120
#SPJ1

Lena [83]1 year ago

4 0

Answer:

(Question) Which quantity is scalar?

(Answer) Energy

(Question) Which option describes a vector quantity?

(Answer) It has magnitude, units, and direction.

(Question) A sailboat ends up 7 miles to the west of a pier. Which statement is true about this situation?

(Answer) It describes displacement, which is a vector quantity.

(Question) Which statement is true regarding a distance vs. time graph?

(Answer) The graph should show distance on the vertical axis.

(Question) On a distance vs. time graph for an object, a line segment that is sloping upward indicates

(Answer) motion in the positive direction.

Explanation:

just finished the quick check UwU

You might be interested in

A proton is moving at 425 m/s. (a) How much work must be done on it to stop it? (A proton has a mass of 1.67×10−27 kg.) (b) Assu

Tpy6a [65]

Answer:

a)1.51*10^-22joules b) 1.89*10^-7m

Explanation:

Work done to stop the proton = the kinetic energy of the proton = 1/2 mv^2 = 1/2* 1.67*10^-27* 425*425 = 1.51* 10 ^ -22 joules

b) net force acting to stop the proton = 8.01*10^-16

Work done needed to stop the proton = net force acting opposite the motion * distance

Distance covered = need work done/ net force

Distance = 1.51*10^-22/8.01*10^-16= 1.89*10^-7m

8 0

3 years ago

You are caulking a window. The caulk is rather thick and, to lay the bead correctly, the exit nozzle is small. A caulking gun us

ZanzabumX [31]

Answer:

The answer is

A. Pressure is distributed uniformly throughout the fluid and the area of the plunger is much larger than the area of the opening.

Explanation:

The question is incomplete, here is a complete question with full options

You are caulking a window. The caulk is rather thick and, to lay the bead correctly, the exit nozzle is small. A caulking gun uses a plunger which is operated by pulling back on a handle. You must squeeze the handle very hard to get the caulk to come out of the narrow opening because:_________.

A. pressure is distributed uniformly throughout the fluid and the area of the plunger is much larger than the area of the opening.

B. viscous drag between the walls of the tip and the caulk causes the caulk to swirl around chaotically.

C. Newton’s third law requires most of the energy in the caulk to be used to push back on the plunger rather than moving it through the tip.

D. the high density of the caulk impedes its flow through the small opening.

Since the caulk is thick and the exit nozzle is small, the pressure needed to deliver the caulk will be very high as pressure is uniformly distributed at the plunger side at every part of the caulk, hence very high pressure is needed to deliver the caulk which is why the handle needed the very hard squeeze

3 0

3 years ago

7. A 1000 kg car is rolling down the street at 2.5 m/s. How fast would a 2500 kg car have to

babunello [35]

1 m/s

Explanation:

To solve this question we use the following formula:

momentum = mass × velocity

momentum of the first car = 1000 kg × 2.5 m/s

momentum of the second car = 2500 kg × X m/s

To bring the cars at rest the momentum of the first car have to be equal to the momentul of the second car.

momentum of the first car = momentum of the second car