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frutty [35]
3 years ago
8

What investigations are best for demonstrating cause and effect relationship

Physics
2 answers:
poizon [28]3 years ago
8 0
The cause would be how the idea started or developed and the effect would be how they both effected you together
ella [17]3 years ago
7 0

Answer:

Experimental investigations are best for demonstrating cause-and-effect relationships.

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PLS ANSWERR THIS QUESTION FOR ME !!
MA_775_DIABLO [31]

Explanation:

20 joule is your answer

Answer:

here

mass m =100kg

distance d=50m

acceleration due to gravity a =10m/s²

work =force×displacement

= ma/d=100×10/50=20joule

3 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP!!!
kirza4 [7]

Answer:

0.5mv^2=50, v=5, 25/2×m=50, m=50×2/25, So, the answer is 4

6 0
3 years ago
What symbol represents energy that moves from a hot object to a cooler object?
strojnjashka [21]
The correct answer is 
<span>C) Q

In fact, the symbol Q represents the heat, which is the form of energy transferred from a hot object to a cooler object. Heat generally refers to the energy related to the motion of the particles, and it is related to the temperature of an object: the higher the temperature of an object, the faster the particles of the object move, and so the object can transfer more energy (as heat) to other objects with lower temperature.</span>
6 0
3 years ago
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,
eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

x(t) = A cos (\omega t + \phi) (1)

where

A is the amplitude

\omega is the angular frequency

\phi is the phase

t is the time

The displacement of the piston in the problem is given by

x(t) = (5.00 cm) cos (5t+\frac{\pi}{5}) (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

v(t) = x'(t) = -\omega A sin (\omega t + \phi) (3)

Differentiating eq.(2), we find

v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})

And substituting t=0, we find the velocity at time t=0:

v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)

Differentiating eq.(3), we find

a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})

And substituting t=0, we find the acceleration at time t=0:

a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

\omega=\frac{2\pi}{T}

Using \omega = 5.0 rad/s and solving for T, we find

T=\frac{2\pi}{5 rad/s}=1.26 s

4 0
3 years ago
Can uh Believe??<br>Am Leaving Coz of it​
zhannawk [14.2K]

Answer:

fo sure

Explanation:

5 0
2 years ago
Read 2 more answers
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