4. Someone took the last cookie in the cookie jar last night. The LAST

What is the magnitude of the speed of a person at the equator due to rotational speed of the Earth?

notka56 [123]

The circumference of the Earth at the equator is listed as 24,901 miles.
So his speed is

   24,901 miles per day.

Convert it to units that we have a better feel for:

   (24,901 mi/da) x (1 da / 24 hrs)

   = (24,901 / 24) (miles/hour)

   = about 1,038 miles per hour.

You'll find a huge number of people on the internet these days,
telling you that you could not be moving at that speed and not
feel it, so therefore the Earth is not spinning, and it's not a globe.

I have a lot of feelings and comments about those people, their
lines of reasoning, and their levels of education and intelligence,
so don't get me started.

I just want to guarantee you that everything you're learning about
the Earth and the solar system in school is well founded, and it's
all based on the life's work of some of the smartest people of the
past 300 years of human history. Everything you're taught about
the Earth has good reasons behind it, whereas those other people
have nothing.

A person on Earth's equator is moving from west to east at roughly
1,038 miles per hour, relative to any point on the Earth's rotation axis.

4 0

3 years ago

Plaease help me with thesse four 40 points

Zielflug [23.3K]

Answer:

For the first one c is the answer

For the second one c is also the answer

For the third one is b

Explanation:

I took that

3 0

2 years ago

The inductance in the drawing has a value of L = 9.4 mH. What is the resonant frequency f0 of this circuit?

yuradex [85]

Answer:

The resonant frequency of this circuit is 1190.91 Hz.

Explanation:

Given that,

Inductance, $L=9.4\ mH=9.4\times 10^{-3}\ H$

Resistance, R = 150 ohms

Capacitance, $C=1.9\ \mu F=1.9\times 10^{-6}\ C$

At resonance, the capacitive reactance is equal to the inductive reactance such that,

$X_C=X_L$

$2\pi f_o L=\dfrac{1}{2\pi f_oC}$

f is the resonant frequency of this circuit

$f_o=\dfrac{1}{2\pi \sqrt{LC}}$

$f_o=\dfrac{1}{2\pi \sqrt{9.4\times 10^{-3}\times 1.9\times 10^{-6}}}$

$f_o=1190.91\ Hz$

So, the resonant frequency of this circuit is 1190.91 Hz. Hence, this is the required solution.

4 0

3 years ago

Which of the following is an example of a safe laboratory produced?

Step2247 [10]

I think you meant to say "procedure" instead of "produced".

Choice B is the answer as it is the safest and most sensible (to prevent damage to clothing or start a fire for example). The other choices are nonsensical or silly.

8 0

3 years ago

Compare these two collisions of a PE student with a wall.

Stolb23 [73]

1) The variable that is different in the two cases is $\Delta t$ , the duration of the collision

2) The change in momentum is the same in the two cases

3) The impulse is the same in the two cases

4) Case B will experience a greater force

Explanation:

1)

The variable that is different in the two cases is $\Delta t$ , the duration of the collision.

In fact, in the first case the wall is padded: this means that the collision will be "softer" and therefore will last longer, so the duration of the collision, $\Delta t$ , will be larger.

In the second case instead, the wall is unpadded: this means that the collision is "harder" and so it will last less time, therefore the duration of the collision $\Delta t$ will be smaller.

2)

The change in momentum in the two cases is the same.

In fact, the change in momentum is given by:

$\Delta p = m(v-u)$

where:

m is the mass of the student

u is the initial velocity

v is the final velocity

In both cases, we have:

m = 75 kg

u = 8 m/s

v = 0 (they both comes to rest)

Therefore, the change in momentum is

$\Delta p = (75)(0-8)=-600 kg m/s$

3)

The impulse in the two cases is the same.

In fact, impulse is defined as the product of force applied, F, and duration of the collision, $\Delta t$ :

$J=F \Delta t$

However, the force can be rewritten as product of mass (m) and acceleration (a), according to Newton's second law:

$F=ma$

So the impulse is

$J=ma\Delta t$

The acceleration can be rewritten as rate of change of velocity:

$a=\frac{\Delta v}{\Delta t}$

So the impulse becomes

$J=m\frac{\Delta v}{\Delta t}\Delta t = m\Delta v$

So, the impulse is equal to the change in momentum: and since in the two cases the change in momentum is the same, the impulse is the same as well.

4)

The force in the collision is related to the impulse by

$J=F\Delta t$

where

J is the impulse

F is the force

$\Delta t$ is the duration of the collision

The equation can be rewritten as

$F=\frac{J}{\Delta t}$

In the two situations described in the problem (A and B), we already said that the impulse is the same (because the change in momentum is the same). However, in case A (padded wall) the time $\Delta t$ is longer, while in case B (unpadded wall) the time $\Delta t$ is shorter: since the force F is inversely proportional to the duration of the collision, this means that in case B the student will experience a greatest force compared to case A.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

3 0

3 years ago