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3 years ago

Mostly question three, but I'm not sure how to work out the speed. Thank you!

Physics

2 answers:

telo118 [61]3 years ago

8 0

The speed of sound follows the same rules, just like any other speed, if the acceleration is zero.

distance = speed * time

distance = 2m
speed = 333m/s

Solve for time t.

Finger [1]3 years ago

4 0

From this question we can obtain the following equestion:
T=displacement ÷ velocity (2m ÷ 333m/s) equal to 6×10^-3

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how would the velocity of the book change if the applied force were equal to the sliding friction force

Inessa [10]

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4 years ago

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The accel

BigorU [14]

Answer:

47.4 m

Explanation:

When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.

In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

$t=\frac{6.22}{2}=3.11 s$

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

$s=ut+\frac{1}{2}gt^2$

where

s is the vertical displacement

u = 0 is the initial velocity

t = 3.11 s is the time

$g=9.8 m/s^2$ is the acceleration of gravity (taking downward as positive direction)

Solving the formula, we find

$s=\frac{1}{2}(9.8)(3.11)^2=47.4 m$

7 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$