A horizontal line above the time axis of a speed vs. time graph means an object is __

A car speed off around a bend at a constant 10m/s explain why it's velocity is not constant

VARVARA [1.3K]

It's velocity is not constant as direction is changing.

We know, velocity is speed with direction, so if direction is changing, velocity can't be constant, doesn't matter that speed is constant.

Hope this helps!

8 0

3 years ago

With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons

Shtirlitz [24]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

$m=\frac{Y-Y_{1}}{X-X_{1}}$ (1)

Where:

$m=5.7$ is the slope of the line

$Y_{1}=2390.7pounds$ is the airplane weight with 51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

$X_{1}=51gallons$ is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

$(X_{1},Y_{1})=(51,2390.7)$

Rewritting (1):

$Y=m(X-X_{1})+Y_{1}$ (2)

As Y is a function of X:

$Y=f_{(X)}=m(X-X_{1})+Y_{1}$ (3)

Substituting the known values:

$f_{(X)}=5.7(X-51)+2390.7$ (4)

$f_{(X)}=5.7X-290.7+2390.7$ (5)

$f_{(X)}=5.7X+2100$ (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

$f_{(81)}=5.7(81)+2100$ (7)

$f_{(81)}=2561.7$ (8) This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

3 0

2 years ago

What are the strengths and limitations of the doppler and transit methods? What kind of planets are easiest to detect with each

Arturiano [62]

$\huge\mathfrak\red{✔Answer:-}$

Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on

Limitations: yield only planet's mass and orbital properties

3 0

3 years ago

Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m

Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

$m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s$