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3 years ago

10

Please help!! The graph in the figure shows the position of a particle as it travels along the x-axis. What is the magnitude of

the average speed of the particle between t = 1.0 s and t = 4.0 s?

1 answer:

Bogdan [553]3 years ago

6 0

Answer:

1.3 m/s

Explanation:

average speed = total distance/ total time

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What distance will a vehicle travel before coming to a complete stop from a speed of 70 mph, (a) When the vehicle is traveling o

OLga [1]

Answer:

(a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be $723.91>SSD>620.2$

(c), The SSD will be 910.5 ft.

Explanation:

Given that,

Speed = 70 mph

Suppose, a perception reaction time of 2.5 sec and the coefficient of friction is 0.35

We need to calculate the stopping sight distance

Using formula of SSD

$SSD=1.47\times v\times t+\dfrac{v^2}{30\times(f\pm g)}$

Where, v = speed of vehicle

t = perception reaction time

f = coefficient of friction

g = gradient of road

(a). If the gradient of road is zero.

Then, the stopping sight distance will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35)}$

$SSD=723.9\ ft$

(b-1). If the gradient of road is 0.1

Then, the stopping sight distance will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35+0.1)}$

$SSD=620.2\ ft$

(b-2). If the grade continuously decrease then the SSD will be increase.

But if the grade is increase then the SSD will be decrease and for flat grade the SSD will be more.

So, The SSD will be $723.91>SSD>620.2$

(c). When the vehicle is traveling downhill on a roadway of constant grade then the vehicle take will be more SSD

So, The SSD will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35-0.1)}$

$SSD=910.5\ ft$

Hence, (a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be $723.91>SSD>620.2$

(c), The SSD will be 910.5 ft.

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Projectile Motion: A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when

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Answer:

Explanation:

The package had the same velocity as the plane when it was dropped. Newton's 1st Law says that "an object in motion tends to stay in motion, at the same velocity, in a straight line unless acted on by an outside force".

There only outside force acting on the package was its weight -- that force is straight down. The horizontal velocity that the plane gave the package continued (as Newton said it would), so as it fell, horizontally it kept pace with the plane.

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Answer:

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Answer:

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