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2 years ago

the student plotted distance v.s time graph in which the line sloped downward. Is the graph correct?

1 answer:

7 0

It could be, depending on WHAT distance was graphed,
and how the student labeled the graph's axes.

Most likely, however, she probably drew the graph ...

-- with time on the x-axis increasing to the right,

-- with distance on the y-axis increasing upward,

-- to show the distance FROM the starting point
NOT the distance remaining to the destination.

If that's how she set it up, or if any two of these items are
the other way around, then the line could not slope downward.
THAT would say that the distance from the starting point is
decreasing as time goes on, or the distance remaining to the
destination is increasing as time goes on. That would be silly.

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Jill is pushing a box across the floor. Which represents the upward force perpendicular to the floor?

Georgia [21]

Answer:

When Jill is pushing a box across a floor, it represents the upward motion and it is natural force is applied.

So it is represented as FN and normal force takes place in considering the force perpendicular to the floor.

It seems to support that forced applied on an object when the object is in contact with other.

Explanation:

5 0

2 years ago

An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period

KatRina [158]

Answer:

The value is $x = 45.99 \ Au$

Explanation:

From the question we are told that

The period of the asteroid is $T = 176 \ years = 176 * 365 * 24 * 60* 60 = 5.55*10^{9}\ s$

Generally the average distance of the asteroid from the sun is mathematically represented as

$R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }$

Here M is the mass of the sun with a value

$M = 1.99*10^{30} \ kg$

G is the gravitational constant with value $G = 6.67 *10^{-11} \ m^3 \cdot kg^{-1} \cdot s^{-2}$

$R = \sqrt[3]{ \frac{6.67 *10^{-11} * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }$

=> $R = 6.88 *10^{12} \ m$

Generally

$1.496* 10^{11} \ m \to 1 Au (Astronomical \ unit )$

$R = 6.88 *10^{12} \ m \ \ \ \ \to \ \ x \ Au$

=> $x = \frac{6.88 *10^{12}}{1.496 *10^{11}}$

=> $x = 45.99 \ Au$

7 0

2 years ago

A unit for measuring frequency is the

jonny [76]

A.) Hertz is the measuring unit for frequency!!
[Your explanation is correct]

Hope this helps!

4 0

3 years ago

A baseball accelerates downward at 9.8m/s. if the gravitational force acting on the baseball is 2.2n what is the baseballs mass

Neko [114]

hope this helps you.....

5 0

2 years ago

At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.5 m/s2. At the same insta

Licemer1 [7]

Answer:

6.96 s

Explanation:

<u>Given:</u>

u = initial speed of the automobile = 0 m/s

a = constant acceleration of the automobile = $2.5\ m/s^2$
v = constant speed of the truck = 8.7 m/s

<u>Assume:</u>

t = time instant at which the automobile overtakes the truck.

At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.

$\therefore \textrm{Distance traveled by the automobile }=\textrm{Distance traveled by the truck}\\\Rightarrow ut+\dfrac{1}{2}at^2=vt\\\Rightarrow (0)t+\dfrac{1}{2}\times 2.5\times t^2=8.7t\\\Rightarrow 1.25t^2=8.7t\\\Rightarrow 1.25t^2-8.7t=0\\\Rightarrow t(1.25t-8.7)=0\\\Rightarrow t = 0\,\,\,or\,\,\, t = \dfrac{8.7}{1.25}\\\Rightarrow t = 0\,\,\,or\,\,\, t = 6.96\\$

Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.

6 0

2 years ago