Which option correctly identifies the graph that represents an object increasing its position with a constant velocity and describes why?

A --It is Graph D because a vertical line indicates the object is increasing its position, and a straight line means it is moving with a constant velocity.

B--It is Graph C because curving upward indicates the object is increasing its position, and a positive slope means it is moving with a constant velocity.

C--It is Graph A because a horizontal line indicates the object is moving with a constant velocity, and the positive value means it is increasing its position.

D--It is Graph B because a straight, diagonal line indicates the object is moving with a constant velocity, and a positive slope means it is increasing its position.

Which Option Correctly Identifies The Graph That Represents An Object Increasing Its Position With A

Answers

Answer 1

Answer: D

Explanation:

Answer 2

The graph that represents an object increasing its position with a constant velocity is Graph B because a straight, diagonal line indicates the object is moving with a constant velocity, and a positive slope means it is increasing its position. Hence, option (D) is correct.

What is velocity?

The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.

Mathematically:

velocity = ( final position - initial position)/time interval

Hence,

Final position = initial position + velocity ×time interval

Hence, the x-t graph of a  object with constant velocity, is linear in nature with a positive slope. Hence, graph (B) is the correct option.

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Related Questions

a child pulls on a string that is attached to a car. if the child does 80.2 J of work while pulling the car 25.0 m, with what force is the child pulling?

Answers

Answer:

F = 3.20 N

Explanation:

Given:

Work done by child = 80.2 j

Distance that the car moves = 25.0 m

We need to find the force acting on the car.

Solution:

Using work done formula as.

[tex]W = F\times d[/tex]

Where:

W = Work done by any object.

F = Force (push or pull)

d = distance that the object moves.

Substitute [tex]W = 80.2\ J\ and\ d =25.0\ m[/tex] in work done formula.

[tex]80.2 = F\times 25[/tex]

[tex]F=\frac{80.2}{25}[/tex]

F = 3.20 N

Therefore, force acting on the car F = 3.20 N

Which of the following is a result of gravitational forces in the Solar System?
A.
the radiation given off by Jupiter
B.
Saturn is further away from the Sun than Earth
C.
the difference in surface temperature on each of the planets
D.
the orbit of moons around their planets in the Solar system

Answers

Answer:

D

Explanation:

the answer is d because gravitational force is what allows them to rotate

hope this was helpful

If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?

Answers

Answer: 2200J

Explanation:

M = 44kg

V = 10m/s

K.E =?

K.E = 1/2MV2 = 1/2 x 44 x (10)^2

K.E = 22 x 100

K.E = 2200J

. A car moves forward up a hill at 12 m/s with a uni-
form backward acceleration of 1.6 m/s2.
a. What is its displacement after 6.0 s?
b. What is its displacement after 9.0 s?

Answers

A) Displacement after 6.0 s 43.2 m uphill.

B) Displacement after 9.0 s 43.2 m uphill.

Explanation:

A car moving upwards in a hill is [tex]12 ms^{-1}[/tex].

Its uniform backward acceleration is [tex]-1.6ms^{-2}[/tex]. (since backward acceleration is a negative acceleration, it is mentioned in negative)

We need to find the displacement of the car after some time.

Using the equation of the motion formula, we know can identify the displacement.

D=[tex]vt+\frac{1}{2} at^2[/tex].

a) Displacement after 6.0 seconds,

D = [tex]12(6.0)+\frac{1}{2}(-1.6)(6.0)^2[/tex].

=[tex]72+\frac{1}{2} (36)(-1.6).[/tex]

=[tex]72+\frac{1}{2}(-57.6).[/tex]

=72-28.8.

D=43.2 m.

b) Displacement after 9.0 seconds,

D= [tex]12(9.0)+\frac{1}{2}(-1.6)(9.0)^2[/tex].

=[tex]108+\frac{1}{2} (81)(-1.6).[/tex]

=[tex]108+\frac{1}{2}(-129.6).[/tex]

= 108-64.8.

D=43.2 m.

The car's displacement after 6.0 s is 14.4 m and after 9.0 s is 43.2 m.

To find the displacement of the car after a given time, we can use the equation:
Displacement (d) = Initial velocity (v) * time (t) + (1/2) * acceleration (a) * time^2

a. After 6.0 s:
Initial velocity (v) = 12 m/s
Acceleration (a) = -1.6 m/s^2 (negative because it's a backward acceleration)
Substituting the values into the equation:
d = (12 m/s) * (6.0 s) + (1/2) * (-1.6 m/s^2) * (6.0 s)^2 = 72 m - 57.6 m = 14.4 m

b. After 9.0 s:
Using the same equation and substituting the new time, we can calculate the displacement:
d = (12 m/s) * (9.0 s) + (1/2) * (-1.6 m/s^2) * (9.0 s)^2 = 108 m - 64.8 m = 43.2 m

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Can somebody please help me with sig figs for this question

Answers

Answer: 760 cW/microgram

2 significant figures

Explanation: solution attached:

Answer:

760 [tex]\frac{cW}{micrograms}[/tex]

Explanation:

You have to know that:

1 gigawatt (GW)=100000000000 centiwatt (cW) 1 kilogram (kg) = 1000000000 micrograms  (μg)

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

So:

[tex]x=\frac{c*b}{a}[/tex]

In this case, you have [tex]\frac{760 GW}{1 kg}[/tex]

In this case, the rule of three is applied as follows: if 1 gigawatt [GW] = 100000000000 centiwatt [cW], 7.600 GW how many cW are they?

[tex]\frac{7.600 GW}{kg} *\frac{100000000000 cW}{1 GW} *\frac{1 kg}{1000000000 micrograms}=760 \frac{cW}{micrograms}[/tex]

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