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Sloan [31]
3 years ago
13

WHICH SHOULD NOT BE PART OF SCIENTIFIC INQUIRY

Physics
1 answer:
umka21 [38]3 years ago
5 0

Answer:

Not collecting the info

Explanation:

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An airplane flies horizontally with a constant speed of 155.0 m/s at an unknown altitude. A package is released out of the airpl
vladimir1956 [14]

Answer:

 y₀ = 1020.3 m

Explanation:

This is a projectile launching exercise, in this case as the package is released its initial vertical velocity is zero.

            y = y₀ + v_{oy} t - ½ g t²

when it reaches the ground its height is zero

           0 = y₀ + 0 - ½ g t²

           y₀ = ½ g t²

           

let's calculate

         y₀ = ½ 9.8 14.43²

         y₀ = 1020.3 m

8 0
3 years ago
A bicyclist travels at the speed of 25 kilometers per hour for 5 hours. How far does the bicyclist go?
mixas84 [53]

Answer:

125 kilometers

Explanation:

Speed= 25km/h

Time = 5 h

Distance= Speed x Time

= 25 x 5

= 125 km

6 0
2 years ago
Read 2 more answers
A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3
den301095 [7]

Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m

h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m

m = 148g*\frac{1kg}{1000g} = 0.148kg

g = 9.8 \frac{m}{s^2}

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J

The increase in potential energy of the ball is 115.82 J

5 0
3 years ago
An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown
ira [324]

Answer:

The specific gravity of the unkown liquid is 15.

Explanation:

Gauge pressure, at the bottom of the tank in this case, can be calculated from

P_{bot}=\gamma_{oil}h_1 + \gamma_{unk}h_2,

where h_1 and h_2 are the height of the column of oil and the unkown liquid, respectively. Writing for \gamma_{unk}, we have

\gamma_{unk} = \frac{P_{bot} - \gamma_{oil}h_1}{h_2} = \frac{65\frac{kN}{m^2} - 8.5\frac{kN}{m^3}\times 5m}{1.5m} = \mathbf{15 \frac{kN}{m^3}}.

Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.

3 0
2 years ago
A student bangs a brick at the head of a table. Three students are positioned equal distance from the head with their hands on t
iris [78.8K]

Explanation:

that the people closer too the head of the table will feel more vibrations than the people at the end of the table. since the vibrations will slow down as they travel farther down the table

Hope this helps!!

5 0
3 years ago
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