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katen-ka-za [31]

2 years ago

9

A rock is tossed straight up from ground level with a speed of 22 m/s. When it returns, it falls into a hole 9.0 m deep. a. What

is the rock’s velocity as it hits the bottom of the hole? b. How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

1 answer:

Vera_Pavlovna [14]2 years ago

5 0

Answer:

(a) $v_{f}=26.08m/s$

(b) $t=4.906seconds$

Explanation:

We will take ground level as origin and upward is positive direction

the givens are

$y_{i}=0\\y_{f}=-9.0m\\ v_{i}=22m/s$

Part (a)

To find the final velocity we use the kinematic equation

So

$(v_{f})^{2} =(v_{i})^{2} +2a(y_{f}-y_{i})\\(v_{f})^{2}=(22m/s)^{2}+2(-9.8m/s^{2} ) (-10-0)\\v_{f}=\sqrt{(22m/s)^{2}+2(-9.8m/s^{2} ) (-10-0)} \\v_{f}=26.08m/s$

Part (b)

To find time of rock trip until it touches the ground we will use simple kinematic equation or simple motion equation

$v_{f}=v_{i}+at\\t=\frac{v_{f}-v_{i}}{a}\\ t=\frac{(-26.08m/s)-(22m/s)}{-9.8m/s^{2} }\\t=4.906seconds$

Notice that we substituted vf with negative sign because its direction is downwards

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Enter a curve slower than the posted speed if your vehicle has a high center of gravity or if surface _____________ is less than

alisha [4.7K]

Answer:

Traction

Explanation:

Enter a curve slower than the posted speed if your vehicle has a high center of gravity or if surface Traction is less than ideal.

Traction is an act of drawing or pulling something over a surface specially a road or a track or it also defined as the grip of a tire on road or a wheel on rail. So speed when entering a curve should be slower when the tire has low traction to avoid accident.

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2 years ago

Nataliya [291]

Answer: b

Explanation: the two pieces will repel as both have obtained a static charge.

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2 years ago

A projectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. It strikes

Alchen [17]

<h2>Answer: x=125m, y=48.308m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=54.5m/s$ is the projectile's initial speed

$\theta=35\°$ is the angle

$t=2.80s$ is the time since the projectile is launched until it strikes the target

$x$ is the final horizontal position of the projectile (the value we want to find)

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the projectile (we are told it was launched at ground level)

$y$ is the final height of the projectile (the value we want to find)

$g=9.8m/s^{2}$ is the acceleration due gravity

Having this clear, let's begin with x (1):

$x=(54.5m/s)cos(35\°)(2.8s)$ (3)

$x=125m$ (4) This is the horizontal final position of the projectile

For y (2):

$y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}$ (5)

$y=48.308m$ (6) This is the vertical final position of the projectile

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2 years ago

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harina [27]

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Explanation:

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7 0

2 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

2 years ago