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frutty [35]
3 years ago
9

Laboratory experiments, observational field studies, and model-building are all examples of different forms of scientific invest

igations. In what way do laboratory experiments primarily differ from other forms of scientific investigations?
A.
A laboratory experiment is the only accepted form of investigation within the scientific community.
B.
Studies about how things behave in nature or studies involving very large objects are best answered through laboratory experiments.
C.
Data can only be generated through laboratory experiments, not other forms of investigation.
D.
Laboratory experiments involve the identification and control of variables.
Physics
1 answer:
Ne4ueva [31]3 years ago
8 0

Answer:

C

Explanation:

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Answer:

<em>a. The rock takes 2.02 seconds to hit the ground</em>

<em>b. The rock lands at 20,2 m from the base of the cliff</em>

Explanation:

Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.

The time taken by the object to hit the ground is calculated by:

\displaystyle t=\sqrt{\frac{2h}{g}}

The range is defined as the maximum horizontal distance traveled by the object and it can be calculated as follows:

\displaystyle d=v.t

The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.

a,

The time taken by the rock to reach the ground is:

\displaystyle t=\sqrt{\frac{2*20}{9.8}}

\displaystyle t=\sqrt{4.0816}

t = 2.02 s

The rock takes 2.02 seconds to hit the ground

b.

The range is calculated now:

\displaystyle d=10\cdot 2.02

d = 20.2 m

The rock lands at 20,2 m from the base of the cliff

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2 years ago
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It's and if, then statement!
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Two balls, each with a mass of 0.890 kg, exert a gravitational force of 8.06 × 10−11 n on each other. how far apart are the ball
Bond [772]
This problem involves Newton's universal law of gravitation and the equation to follow would be.

F = GM₁M₂/r²  

Given: M₁ = 0.890 Kg;  M₂ = 0.890 Kg;  F = 8.06 x 10⁻¹¹ N; G = 6.673 X 10⁻¹¹ N m²/Kg²

Solving for distance r = ?

r = √GM₁M₂/F

r = √(6.673 x 10⁻¹¹ N m₂/Kg²)(0.890 Kg)(0.890 Kg)/ 8.06 x 10⁻¹¹ N

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A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at
Vesna [10]

Answer:

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0  So B=0

-16 B-32 A+16B=6  So A=-3/16

y=-3/16 cos 4t

Now to find the CF  of differential equation 1

y"+8 y'+ 16y=6 cos 4 t

Homogeneous version of above equation

m^2+8m+16=0

So CF =(C_1+tC_2)e^{-2t}

So the general equation

Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t

Given that t=0 Y=0 So

C_1=\dfrac{3}{16}

t=0 Y'=0 So

C_2 =\dfrac{3}{8}

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

The above equation is the general equation for motion.

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