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2 years ago

Currently, Earth's tilt is closest to _____.

1 answer:

4 0

The answer is C. 23.5°. It's because of this tilt that the Earth experiences seasons as it orbits around the Sun. Imagine the Sun is at the center of a spinning record.

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As a stands, her entire weight is momentarily placed on the heels of her high-heeled shoes. Calculate the pressure exerted on th

Lena [83]

Answer:

Pressure will be $6072449.952Pa$

Explanation:

We have given mass of the women m = 65 kg

Radius of the heels r = 0.578 cm = 0.00578 m

We have to find the pressure

We know that pressure is given by

$P=\frac{F}{A}=\frac{mg}{A}$

So force F = mg = 65×9.8 = 637 N

Area $A=\pi r^2=3.14\times 0.00578^2=1.049\times 10^{-4}m^2$

So pressure $p=\frac{637}{1.049\times 10^{-4}}=6072449.952Pa$

5 0

3 years ago

What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a

Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

v = r ω

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}$

$I = \dfrac{m(2gh - v^2)r^2}{v^2}$

$I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}$

$I =mr^2(0.287)$

I = 0.287 MR²

3 0

2 years ago

What is the thermal energy of an object?

mr Goodwill [35]

The thermal energy of an object is the energy contained in the motion and vibration of its molecules. Thermal energy is measured through temperature. The energy contained in the small motions of the object's molecules can be broken up into a combination of microscopic kinetic energy and potential energy.

5 0

2 years ago

Can a plane mirror produce a real image? Explain.

Komok [63]

I don't think so, because in order to produce an image, you need a surface behind the mirror. The light will hit the mirror, then it will bounce it back in your eyes and you see the image.

4 0

2 years ago

A 1000-kg car traveling at 70 m/s takes 3 m to stop under full braking. the same car under similar road conditions, traveling at

azamat

We assume $a=const$ (acceleration is constant. We apply the equation
$v^2=v0^2+2as$ where s is the distance to stop $v=0(m/s)$ . We find the acceleration from this equation
$a=-v0^2/(2s)=-70^2/(2*3) =-816.7 (m/s^2)
$
We know the acceleration, thus we find the distance necesssary to stop when initial speed is $v=140 (m/s)$
$s=-v0^2/(2a) =140^2/(2*816.7)=12 (m)$

5 0

3 years ago