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3 years ago

Which best describes the scientific method?

Physics

2 answers:

EleoNora [17]3 years ago

6 0

The answer on ED:

D: "the process of hypothesis and testing through which scientific inquiry occurs"

Mrrafil [7]3 years ago

3 0

Your answer would be, The process of Hypothesis, and Testing through which scientific inquiry occurs.

Hope that helps!!!! : )

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Which would deliver a greater change in momentum to an opponent’s body - a dodgeball that traveled at 10m/s and rebounded with a

Mazyrski [523]

2nd sentence seems bout right ?? i think

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3 years ago

A series circuit contains a 9-volt battery, a 3-ohm resistor and a 2-ohm resistor. What is the voltage drop across the 2-ohm res

BARSIC [14]

Since everything in the circuit is in series .. .

-- The total resistance is (3 + 2) = 5 ohms.

-- The voltage across the 3-ohm resistor is 3/5 of the total voltage.

-- The voltage across the 2-ohm resistor is 2/5 of the total voltage.

(2/5) of (9 volts) = 18/5 = 3.6 volts .

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3 years ago

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3 years ago

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PLEASE HELPPPPP!!!! :)

GuDViN [60]

Answer:

F=X.F=mxq. 3. 1 N/kg=0.5kg g=9.80

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2 years ago

The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.

Salsk061 [2.6K]

Answer:

$5.94\ \text{m/s}$

$1.7$

$0.577$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$\theta$ = Angle of slope = $30^{\circ}$

v = Velocity of child at the bottom of the slide

$\mu_k$ = Coefficient of kinetic friction

$\mu_s$ = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}$

The speed of the child at the bottom of the slide is $5.94\ \text{m/s}$

Length of the slide is given by

$l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}$

$v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}$

The force energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73$

The coefficient of kinetic friction is $1.7$ .

For static friction

$\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577$

So, the minimum possible value for the coefficient of static friction is $0.577$ .

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2 years ago