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2 years ago

10

If a cliff jumper leaps off the edge of a 100m cliff, how long does she fall before hitting the water? (assume zero air resistan

ce)

1 answer:

andrew-mc [135]2 years ago

4 0

<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

Assuming that the initial velocity of the jumper is zero, on Earth any freely falling object has an acceleration of 9.8 m/s².

<em>✔ We have : a = v/Δt = ⇔ Δt = v/a </em>

Δt = (√2xgxh)/9,8
Δt = (14√10)/9,8
Δt ≈ 4,5 s

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Under which condition would time periods of daylight and darkness be equal everywhere on Earth all year? A. if Earth revolved ar

Marizza181 [45]

The correct answer is
<span>B. if Earth rotated on an axis that was not tilted with respect to Earth's orbit

In fact, the fact that Earth's axis is tilted is the reason why the durations of day and night are different in every part of the Earth. If the axis was not tilted, we would have exactly 12 hours of day and night in every point of the Earth, for the whole year.</span>

7 0

3 years ago

Read 2 more answers

one car accelerates at half the rate of another how much longer does it take the first car to travel a quarter mile

Orlov [11]

Answer:

t=1/4v1

Explanation:

Given data

Car one

Speed =v1

Time =t

Distance =1/4 mile

Given data

Car two

Speed =v1/2

Time =t

Distance =d

Speed =distance/time

v1=1/4/t

v1t=1/4

t=1/4*1/v1

t=1/4v1 seconds

7 0

2 years ago

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

3 years ago

Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1

miv72 [106K]

Answer:

w = 4,786 rad / s , f = 0.76176 Hz

Explanation:

For this problem let's use the concept of angular momentum

L = I w

The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved

Initial Before sticking

L₀ = 0 + I₂ w₂

Final after coupling

$L_{f}$ = (I₁ + I₂) w

The moments of inertia of a disk with an axis of rotation in its center are

I = ½ M R²

How the moment is preserved

L₀ = $L_{f}$

I₂ w₂ = (I₁ + I₂) w

w = w₂ I₂ / (I₁ + I₂)

Let's reduce the units to the SI System

d₁ = 60 cm = 0.60 m

d₂ = 40 cm = 0.40 m

f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz

Angular velocity and frequency are related.

w₂ = 2 π f₂

w₂ = 2π 3.33

w₂ = 20.94 rad / s

Let's replace

w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)

w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

Let's calculate

w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)

w = 20.94 1.28 / 5.6

w = 4,786 rad / s

Angular velocity and frequency are related.

w = 2π f

f = w / 2π

f = 4.786 / 2π

f = 0.76176 Hz

7 0

3 years ago

Which of the following statements is true about earth's magnetic poles

Dahasolnce [82]

Answer: b) they are the areas where Earth's magnetic field is weakest

Explanation:

According to classical physics, a magnetic field always has two associated magnetic poles (north and south), the same happens with magnets. This is because for <em>classical physics</em>, naturally, magnetic monopoles can not exist.

In this context, Earth is similar to a magnetic bar with a north pole and a south pole. This means, the axis that crosses the Earth from pole to pole is like a big magnet.

Now, by convention, on all magnets the north pole is where the magnetic lines of force leave the magnet and the south pole is where the magnetic lines of force enter the magnet. Then, for the case of the Earth, the north pole of the magnet is located towards the geographic south pole and the south pole of the magnet is near the geographic north pole.

Being the magnetic poles the places where the Earth's magnetic field is weakest. And it is for this reason, moreover, that the magnetic field lines enter the Earth through its magnetic south pole (which is the geographic north pole).

4 0

3 years ago