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VARVARA [1.3K]
3 years ago
6

How are scientific theories different than laws or hypotheses?

Physics
2 answers:
Veronika [31]3 years ago
7 0

Answer:

A

Explanation:

Hope this helps you

Nina [5.8K]3 years ago
3 0

Answer:The answer is A

Explanation:

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Psychophysics is a discipline in psychology that focuses solely on the identification of stimuli. T F
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This is False.
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How to calculate the slope of position time graph?
Mashcka [7]

Answer:

The slope of a position-time graph can be calculated as:

m=\frac{\Delta y}{\Delta x}

where

\Delta y is the increment in the y-variable

\Delta x is the increment in the x-variable

We can verify that the slope of this graph is actually equal to the velocity. In fact:

\Delta y corresponds to the change in position, so it is the displacement, \Delta s

\Delta x corresponds to the change in time \Delta t, so the time interval

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m=\frac{\Delta s}{\Delta t}

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3 0
4 years ago
PLEASE HELP
Elan Coil [88]

Answer:

Explanation:

a_{x}=0 v_{xo}= v_{o}cos(delta)t a_{y}=-g v_{yo}=sinθ

X-direction                                     | Y-direction

x=x_o+v_{xo}t ⇒ x=v_{xo}cos(delta)t |

5 0
3 years ago
A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.90 m/s2 . At 20.0s after blastoff, t
Brilliant_brown [7]

Answer:

A) 580m

B) 0 m/s

C) 9.8m/s^2

D) downward

E) 10.87s

F) 106.62 m/s

Explanation:

A) The distance traveled by the rocket is calculated by using the following expression:

y=\frac{1}{2}at^2

a: acceleration of the rocket = 2.90 m/s^2

t: time of the flight = 20.0 s

y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m

B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.

C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2

D) The acceleration points downward

E) The time the rocket takes to return to the ground is given by:

t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s

10.87 seconds

F) The velocity just before the rocket arrives to the ground is:

v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}

6 0
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