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4 years ago

Which will most likely happen on the international space station?

Physics

2 answers:

r-ruslan [8.4K]4 years ago

8 0

B. A safe landing on the station on earth
hope this helps :D

monitta4 years ago

8 0

The answer is D. An Experiment To Study The Growth Of Cells In Space. Hope this helps!! <3

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Mekhanik [1.2K]

Answer:

yes h h jj h v gu i kj

Explanation:

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3 years ago

Under what atmospheric condition is fog most often formed in the san joaquin valley?

GalinKa [24]

It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway,here is the answer. The atmospheric condition in which fog is most often formed in the san Joaquin valley is stable stability. Hope this helps.

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4 years ago

Help pls anybody pro at this

aniked [119]

Answer:

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3 years ago

Read 2 more answers

12*3a car is stopped at a traffic light. it then travels along a straight road so that its distance from the light is given by w

Digiron [165]

Missing parts in the text of the exercise:
- The distance from the traffic light is given by $x(t) = bt^2 -ct^3$ , where $b=2.4 m/s^2$ and $c=0.13 m/s^3$

Solution:

part a) calculate the average velocity of the car for the time interval t=0 to t=8.0 s
- The average velocity is given by the ratio between the distance covered in the time interval:
 $v_{ave} = \frac{x(8.0 s)-x(0 s)}{8.0 s-0 s}$
x(0 s), the distance covered after t=0 s, is zero, while the distance after t=8.0 s is
$x(8.0 s)=b(8 s)^2-c(8 s)^3 = (2.4 m/s^2)(8 s)^2 -(0.13 m/s^3)(8 s)^3=$
$=87.04 m$
Therefore, the average velocity is
$v_{ave}= \frac{x(8.0 s)-x(0)}{8.0s-0}= \frac{87.04 m}{8.0s}= 10.88 m/s$

part b) calculate the instantaneous velocity of the car at t=0, t=4.0 s and t=8.0 s
- The instantaneous velocity can be found by performing the derivation of x(t):
 $v(t) = \frac{dx(t)}{dt}=2bt-3ct^2$
So now we just have to substitute t=0, t=4 s and t=8 s:
- t=0: v(0)=0
- t=4 s: $v(4.0 s)=2(2.4 m/s^2)(4.0 s)-3(0.13 m/s^2)(4.0s)^2=12.96 m/s$
- t=8 s: $v(8.0 s)=2(2.4 m/s^2)(8.0 s)-3(0.13 m/s^2)(8.0s)^2=13.44 m/s$

part c) how long after starting from rest is the car again at rest?
- To solve this part we must find the value of t for which v(t)=0, so:
 $2bt-3ct^2=0$
$t(2b-3ct)=0$
The first solution is t=0 s, which corresponds to the beginning of the motion, so we are not interested in this value. The second solution is
 $t= \frac{2b}{3c}= \frac{2\cdot 2.4 m/s^2}{3 \cdot 0.13 m/s^3}=12.31 s$
and this is the time at which the car is at rest again.

8 0

4 years ago

How much energy will a stock tank heater rated at 1790 Watts use in a 24 hour period? 1. 1790 × 24 × 3600 Joules 2. 1790 × 24 ×

Rashid [163]

<h2>Option 1 is the correct answer.</h2>

Explanation:

Power of heater, P = 1790 W

Time used, t = 24 hours = 24 x 60 x 60 = 24 x 3600 s

We have the equation

$\texttt{Power}=\frac{\texttt{Energy}}{\texttt{Time}}$

We need to find energy,

Substituting

$\texttt{Power}=\frac{\texttt{Energy}}{\texttt{Time}}\\\\1790=\frac{\texttt{Energy}}{24\times 3600}$

Energy = 1790 x 24 x 3600 J

Option 1 is the correct answer.

3 0

3 years ago