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3 years ago

11

What is the linear speed of a point (a) on the equator, (b) on theArctic Circle (latitude 66.5o N), and (c) at a latitudeof 45.0

degrees N, due to the Earth's rotation?
For part (a), I used the equation w = A/t to represent that theEarth rotates once a day on its axis. A point on the equatorthus rotates through 2πradians in 24 hours, so
w = A/t = (2π) / (24 hr) = (6.28 rad) / (24hr x 3600 s/hr) = 7.27 x 10-5 rad/s
Since the radian is a unitless measure it is not necessary toinclude it in the units for a quantity.
The radius of the Earth is re = 6.4 x 106 m,so:
v = wr = (7.27 x 10-5 rad/s)(6.4 x106 m) = 465 m/s = 1674 km/hr

1 answer:

KiRa [710]3 years ago

8 0

Answer:

a) v = 465.3 m / s , b) v = 196.3 m / s , c) v =327.2 m/s

Explanation:

Linear and angular magnitudes are related.

v = w r

Where v and w are the linear and angular velocities, respectively and r is the radius of the circle.

Let's look for angular velocity

In rotated angle is a rotation of the Earth is 2π rad and the Time spent Period is24 h

Let's reduce the hours to seconds

T = 24 h (3600 s / 1 h) = 86,400 s

w = 2π / 68400

w = 7.27 10⁻⁵ rad / s

Let's look for linear speeds

a) in Ecuador the radius of the circle is the radius of the planet

r = $R_{e}$ = 6.4 10⁶ m

v = 7.27 10⁻⁵ 6.4 10⁶

v = 465.3 m / s

b) As the sphere reduces the size of the circle, let's use trigonometry to find the triangle leg, the angle is measured from the x-axis (East

cos 65 = r / $R_{e}$

r = $R_{e}$ cos 65

r = 6.4 10⁶ cos 65

r = 2.7 10⁶ m

We calculate the linear speed

v = 7.27 10⁻⁵ 2.7 10⁶

v = 196.3 m / s

c) at an angle of 45

r = 6.4 10⁶ cos45

r = 4.5 10⁶ m

v = 7.27 10⁻⁵ 4.5 10⁶

v =327.2 m/s

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Answer:

MRCORRECT has answered the question

Explanation:

Since velocity is a vector, it can change either in magnitude or in direction. Acceleration is therefore a change in either speed ordirection, or both. Keep in mind that althoughacceleration is in the direction of the changein velocity, it is not always in the direction ofmotion.

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An empty 2,500 kg train car is headed northbound at a velocity of 5 m/s. Ahead of the first car, an empty 1,500 kg car is headed

Tema [17]

Let the mass of 2500 kg car be $m_1$ and it's velocity be $v_1$ and the mass of 1500 kg car be $m_2$ and it's velocity be $v_2$ .

After the bumping the mass be M and it's velocity be V.

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So the velocity of the two-car train = 3.5 m/s

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3 years ago

An electric pump rated 1.5 KW lifts 200kg of water through a vertical height of 6m in 10 secs: way is the efficiency of the pump

ruslelena [56]

Answer:

80%

Explanation:

Efficiency = Power output / Power input × 100 %

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3 years ago

If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is

viktelen [127]

To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

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3 years ago

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Marysya12 [62]

Answer:

$v_{mp}=305.83 m/s$

Explanation:

The temperature in stratosphere is generally about 270 K

molecular weight of an ozone molecule = 48 gm/mole

now formula for most probable velocity

$v_{mp}= \sqrt{\frac{2RT}{M} }$

plugging the values we get

$v_{mp}= \sqrt{\frac{2\8.314\times270}{48} }$

$v_{mp}=305.83 m/s$

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3 years ago