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Brums [2.3K]
3 years ago
9

What is everyone's take on time travel?

Physics
1 answer:
Andrew [12]3 years ago
7 0

Answer:

Its not really possible I don't think. UNLESS! You fall into a manhole then find a wirling vortex in the sewers! : )

Explanation:

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Help please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Tom [10]

A2 = 16 m^2

Explanation:

Application of Pascal's law:

F1/A1 = F2/A2

Given:

F1 = 50 N. A1 = 1 m^2

F2 = 800 N A2 = ?

A2 = (F2/F1)A1 = (800 N/50 N)(1 m^2)

= 16 m^2

5 0
3 years ago
Two similar gold spheres are separated by a distance of
solong [7]

Answer:

The answer to your question is the letter D. 2.5 N

Explanation:

The electrostatic force is the same in both directions,

If the electrostatic force on B due to A is 2.5 N, the magnitude of the electrostatic force on A due to B must be 2.5N.

Maybe the direction is different but the magnitude is the same.

5 0
3 years ago
Explain what is meant by “In Phase” and “Out of Phase?
LekaFEV [45]
In phase would mean both waves are at a positive peak, out of phase would mean one is at a positive whilst the other is at a negative. Out of phase would mean the waves cancel each other out
5 0
3 years ago
Read 2 more answers
Find the magnitude of the sum of these two vectors: B 101 m 60.0 ° 85.0 m A​
slamgirl [31]

Answer:

M= 161.1 m Deg= 32.9

Explanation:

3 0
2 years ago
a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik
grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:  

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)  

Where:  

y=0 is the height of the ball when it hits the ground  

y_{o}=70 m is the initial height of the ball

V_{o}=82m/s is the initial velocity of the ball  

t is the time when the ball strikes the ground

g=9.8m/s^{2} is the acceleration due to gravity  

Having this clear, let's find t from (1):  

0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2} (2)  

Rewritting (2):

-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0 (3)  

This is a quadratic equation (also called equation of the second degree) of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}  (4)

Where:

a=-\frac{1}{2}(9.8m/s^{2}

b=82m/s

c=70m

Substituting the known values:

t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}  (5)

Solving (5) we find the positive result is:

t=17.68 s

7 0
3 years ago
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