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Damm [24]
3 years ago
7

You want to arrive at your friend's house by 5pm her house is 240 kilometers away if your average speed will be 80km on the trip

when do you need to leave your house in order to get to her house in time
Physics
1 answer:
Annette [7]3 years ago
3 0
You must leave your house by 2:00 pm. 
what you do is you take your total distance and divide that by speed so in this case it is 240km / 80kmh = 3 hours meaning you have to leave the house by 2:00 pm
 
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La velocidad de la luz en el vacío es c= 3000.000 km\s la luz del sol tarda en llegar a la tierra 8 minutos y 14 segundos
ehidna [41]

La velocidad correcta de la luz en el vacío es  300.000 km/s .

La distancia = (velocidad) x (duración de tiempo)

Duración de tiempo = 494 segundos, porque cada minuto = 60 segundos

La distancia = (300.000 km/s) x (494 s)

<em>La distancia = 148.200.000 km</em>

3 0
3 years ago
A banana peel has lots of friction.<br> True or False
olga_2 [115]

Answer:

False

Explanation:

I learned it the hard way trust me T^T

3 0
3 years ago
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
7nadin3 [17]

Answer:

(a) r = 1.062·R_E = \frac{531}{500} R_E

(b) r = \frac{33}{25} R_E

(c) Zero

Explanation:

Here we have escape velocity v_e given by

v_e =\sqrt{\frac{2GM}{R_E} } and the maximum height given by

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}

Therefore, when the initial speed is 0.241v_e we have

v = 0.241\times \sqrt{\frac{2GM}{R_E} } so that;

v² = 0.058081\times {\frac{2GM}{R_E} }

v² = {\frac{0.116162\times GM}{R_E} }

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r} is then

\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}

Which gives

-\frac{0.941919}{R_E} = -\frac{1}{r} or

r = 1.062·R_E

(b) Here we have

K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m  \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}

Therefore we put  \frac{0.241GM}{R_E} in the maximum height equation to get

\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}

From which we get

r = 1.32·R_E

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0
3 years ago
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sasho [114]
Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
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v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around
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timofeeve [1]
The correct answer is "C". 'Old theories are adjusted to incorporate all old new information.' This makes the most sense, regarded the old and new information should be taken into consideration.

I hope this helped you!

Brainliest answer is always appreciated!
8 0
3 years ago
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