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3 years ago

10

Wil-E-Coyote drops a bowling ball off a cliff to try to catch the Roadrunner. The cliff is 132m high. how far does it fall in th

e first 3.0 seconds

1 answer:

Ksivusya [100]3 years ago

4 0

197m because 132+3.0=197x

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Need help with question 1) and 2) please ASAP ??? 15 points ?

Vaselesa [24]

1- interaction between 2 objects
2- action- reaction force pairs

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3 years ago

How long did it take Fernando to drive to

barxatty [35]

How do you calculate distance over speed?
Image result for if your going 30 m/s to go somewhere thats 1680 miles away
The formula can be rearranged in three ways:
speed = distance ÷ time.
distance = speed × time.
time = distance ÷ speed.

1680÷30 = 56

So it would take around 56 minutes to get to Kroger.

I hope this helps !! :)

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1 year ago

Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

<h3>a.</h3>

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

so

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

<h3>b.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

<h3>c.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

5 0

3 years ago

The movement of a magnetic pole away from the actual pole

Lunna [17]

The North Magnetic Pole is the point on the surface of Earth's Northern Hemisphere at which the planet's magnetic field points vertically downwards (in other words, if a magnetic compass needle is allowed to rotate about a horizontal axis, it will point straight down). There is only one location where this occurs, near (but distinct from) the Geographic North Pole and the Geomagnetic North Pole.

3 0

3 years ago

An experiment is conducted using 2 plants to determine if the amount of sunlight they receive affects how fast they grow. Which

Phantasy [73]

Greetings!

The correct answer choice is Choice 4.

<em>Why?</em>

In a scientific experiment the only thing being changed is the independent variable. Everything else should stay the same.

In this experiment, the independent variable is the amount of sunlight each plant should receive. <em>Here's a tip</em>- when looking for and independent variable look for whats being changed on purpose.

Hope this helps!

~Fluerie

4 0

3 years ago