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10 months ago

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

2 answers:

Nadusha1986 [10]10 months 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.

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Lena [83]10 months 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:

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2 years ago

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

Now the horizontal component of velocity:

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

The vertical component of the velocity:

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

Time taken by the projectile to travel the distance of 30 m:

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

Vertical position of the projectile at this time:

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

Now this height should be the maximum height of the tossed apple where its velocity becomes zero.

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

Time taken to reach this height:

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

We observe that $t>t'$ hence the time after the launch of the projectile after which the apple should be tossed is:

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

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3 years ago

If the change in velocity increases, what happens to the acceleration during the same time period?

4vir4ik [10]

When the velocity is increasing the acceleration increases too

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3 years ago

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