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4 years ago

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A girl at an airport rolls a ball north on a moving walking that moves east. If the ball’s speed with respect to the walkway is

0.15m/s and the walkway moves at a speed of 1.50m/s, what is the velocity of the ball relative to the ground?

1 answer:

sweet-ann [11.9K]4 years ago

4 0

We have to add two vectors.

Vector #1: 0.15 m/s north

Vector #2: 1.50 m/s east

Their sum:

Magnitude: √(0.15² + 1.50²)

Magnitude = √(0.0225+2.25)

Magnitude = √2.2725

Magnitude = <em>1.5075 m/s</em>

Direction = arctan(0.15/1.50) north of east

Direction = <em>5.71° north of east</em>

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Answer:

Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s

Explanation:

Given Data:

V is the overall velocity vector, $Vi$ and $Ui$ are its initial vertical and horizontal components

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To find:

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Calculation:

1) Using the $3^{rd}$ equation of motion, we know

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putting values

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4 years ago

The earth is composed of four layers three of these layers are solid and only one is liquid which layer exits in the liquid stat

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A figure skater skates across a rink of length 50 m in 12.1 seconds. a. What is the average speed of the skater? (2 points) b. I

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(a) The skater covers a distance of S=50 m in a time of t=12.1 s, so its average speed is the ratio between the distance covered and the time taken:
$v= \frac{S}{t}= \frac{50 m}{12.1 s}=4.13 m/s$

(b) The initial speed of the skater is
$v_i = 4 m/s$
while the final speed is
$v_f = 5.3 m/s$
and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by
$a= \frac{v_f - v_i}{t}= \frac{5.3 m/s-4.0 m/s}{2.0 s}=0.65 m/s^2$

(c) The initial speed of the skater is
$v_i = 13.0 m/s$
while the final speed is
$v_f=0$
since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:
$2aS=v_f^2 -v_i^2$
from which we find
$a= \frac{-v_i^2}{2S}= \frac{-(13.0m/s)^2}{2 \cdot 8.0 m}=-10.6 m/s^2$
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4 years ago

Hurry Please !!!!!<br><br> Study the diagram<br> Point C identifies the____<br> of the wave

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3 years ago

A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

Scorpion4ik [409]

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where

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Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

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$T=\frac{1}{0.42 Hz}=2.38 s$

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The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

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We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

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3 years ago

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