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podryga [215]
3 years ago
13

Now consider a different time interval: the interval between the initial kick and the moment when the ball reaches its highest p

oint. We want to find how long it takes for the ball to reach this point, and how high the ball goes. (e) What is the component of the ball’s velocity at the instant when the ball reaches its highest point (the end of this time interval)?
Physics
1 answer:
scZoUnD [109]3 years ago
3 0

Answer:A ball is kicked from a location < 9, 0, -8 > (on the ground) with initial velocity < -11, 18, -5 > m/s. The ball's speed is low enough that air resistance is negligible. What is the velocity of the ball 0.5 seconds after being kicked? (Use the Momentum Principle!) = m/s In this situation (constant force), which velocity will give the most accurate value for the location of the ball 0.5 seconds after it is kicked? The arithmetic average of the initial and final velocities. The final velocity of the ball. The initial velocity of the ball. What is the average velocity of the ball over this time interval? avg = Use the average velocity to find the location of the ball 0.5 seconds after being kicked: = m Now consider a different time interval: the interval between the initial kick and the moment when the ball reaches its highest point. We want to find how long it takes for the ball to reach this point, and how high the ball goes. What is the y-component of the ball's velocity at the instant when the ball reaches its highest point (the end of this time interval)? vyf = m/s. Fill in the missing numbers in the equation below (update form of the Momentum Principle): mvyf = mvyi + Fnet,y?t m = m + ?mg?t How long does it take for the ball to reach its highest point? ?t = s. Knowing this time, first find the y-component of the average velocity during this time interval, then use it to find the maximum height attained by the ball: ymax = m. Now take a moment to reflect on the reasoning used to solve this problem. You should be able to do a similar problem on your own, without prompting. Note that the only equations needed were the Momentum Principle and the expression for the arithmetic average velocity.

Explanation:

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A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w
frosja888 [35]

Answer:

(a) n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) n_{2} = 1.349

(c) v_{1} = 2.04\times 10^{8}\ m/s

(d) v_{2} = 2.22\times 10^{8}\ m/s

Solution:

As per the question:

Refractive index of medium 1, n_{1} = 1.47

Angle of refraction for medium 1, \theta_{1} = 59^{\circ}

Angle of refraction for medium 2, \theta_{1} = 69^{\circ}

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

n_{1}sin\theta_{1} = n_{2}sin\theta_{2}

where

n_{2} = Refractive Index of medium 2

Now,

n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}

n_{2} = 1.349

(c) To calculate the velocity of light in medium 1:

We know that:

Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}

Thus for medium 1

n_{1} = \frac{c}{v_{1}

v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s

(d) To calculate the velocity of light in medium 2:

For medium 2:

n_{2} = \frac{c}{v_{2}

v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s

5 0
3 years ago
Read 2 more answers
The physical phenomenon that MRI is based on does not depend on ionizing radiation, but on other properties of atoms instead. Wh
hammer [34]

This phenomenon is called Nuclear Magnetic Resonance.

When I met my future wife, she was working in the medical research building next door to the communications building where I worked. (We shared a parking lot.) MRI was not a thing yet, and she was doing research in Nuclear Magnetic Resonance. I learned a lot about it when I walked next door to visit her in her lab. Strange as it may seem, several years earlier, her older brother was involved in the invention of the CAT scan. When we got married, I figured that our kids had at least a 50% chance of inheriting some brains. So we had some, and they've done OK.

4 0
3 years ago
Read 2 more answers
Un camion de envios se encuentra detenido en una señal de pare, permitiendo que pase una ambulancia. Inicia su recorrido y al ca
Nesterboy [21]

Answer:

0.741\ \text{m/s}^2

Explanation:

v = Velocidad final = 40\ \text{km/h}=\dfrac{40}{3.6}\ \text{m/s}

u = Velocidad inicial = 0

t = Tiempo empleado = 15 s

a = Aceleración

De las ecuaciones cinemáticas tenemos

v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{40}{3.6}-0}{15}\\\Rightarrow a=0.741\ \text{m/s}^2

La aceleración del camión en el primer intervalo de tiempo es 0.741\ \text{m/s}^2.

4 0
3 years ago
A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

  • Mass of the first puck = m_1\ =\ 5\ kg
  • Mass of the second puck = m_2\ =\ 3\ kg
  • initial velocity of the first puck = u_1\ =\ 3\ m/s.
  • Initial velocity of the second puck = u_2\ =\ -1.5\ m/s.

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = v_2\ =\ 2.31\ m/s.

Let v_1 be the final velocity of first puck after the collision.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = 25\times 10^{-3}\ sec

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0
3 years ago
QUICK WILL MARK BRAINLIEST
Tju [1.3M]

Answer:

37.1°

Explanation:

4 0
3 years ago
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