What is a primary source?

If the position of a particle on the x-axis at time t is âˆ’5t2, then the average velocity of the particle for 0 â‰¤ t â‰¤ 3 is

Drupady [299]

Answer:

v = 15 m / s

Explanation:

In this exercise we are given the position function

x = 5 t²

and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval

$v= \frac{v_{f} - v_{o} }{t_{f} - t_{o} }$

let's look for the displacements

t = 0 x₀ = 0 m

t = 3 $x_{f}$ = 5 3 2

x_{f} = 45 m

we substitute

$v = \frac{45 -0}{3 - 0}$

v = 15 m / s

3 0

3 years ago

My school doesn’t really teach help

Olegator [25]

C
Unbalanced causes something to move because the net force is greater than zero

7 0

3 years ago

A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre

pantera1 [17]

Answer:

Part a)

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

t = 12 s

Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

$T = \frac{m}{L}xg + Mg$

so the speed of the wave at that position is given as

$v = \sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{m}{L}$

now we have

$v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}$

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

time taken by the wave to reach the top is given as

$t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}$

$t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})$

$t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})$

$t = 12 s$

4 0

3 years ago

If

nadya68 [22]

Answer:

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Explanation:

hiiii look forward but I don't know how to do it

8 0

3 years ago

A boy can swim 3.0 meter a second in still water while trying to swim directly across a river from west to east, he is pulled by

lana66690 [7]

Answer:

Angle: $48.19^o$

Explanation:

<u>Two-Dimension Motion</u>

When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.

Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.

To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

$v_b\ cos\alpha$

where $v_b$ is the speed of the boy in still water and $\alpha$ is the angle respect to the shoreline. If the river flows at speed $v_s$ , we now set

$v_b\ cos\alpha=v_s$

$\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}$

$\alpha=48.19^o$

8 0

3 years ago