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Musya8
1 month ago
12

Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.7 m tall window in 0.129 sec. From how high above th

e top of the window was the rock dropped? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.
Physics
1 answer:
ValentinkaMS [3.4K]1 month ago
8 0
h = 22.35 m.
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What is the energy equivalent of an object with a mass of 2.5 kg? 5.5 × 108 J 7.5 × 108 J 3.6 × 1016 J 2.25 × 1017 J
Maru [3345]
Utilizing Einstein's equation that signifies the relationship between mass and energy:
E=mc^2
where
E stands for energy
m represents the object's rest mass
and c denotes the speed of light

By applying this formula and using m=2.5 kg, we can compute the object's equivalent energy:
E=mc^2 = (2.5 kg)(3 \cdot 10^8 m/s)^2=2.25 \cdot 10^{17} J
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If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
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Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

Consequently, the temperature difference across the material will be \Delta T = 375 K

Explanation:

In this case, we apply the Fourier Law of heat conduction expressed by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 denotes the cross-sectional area

Q= 3KW signifies the heat transfer rate

\Delta T is the temperature difference we need to determine

represents the thickness of the material\Delta x=2.5 cm =0.025 m

To isolate \Delta T from equation (1), we obtain:

\Delta T =\frac{Q \Delta x}{Ak}

Initially, we convert 3KW to W, resulting in:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

With all variables accounted for, we can substitute and calculate:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

Thus, the temperature difference across the material will be \Delta T = 375 K

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An object of mass 8.0 kg is attached to an ideal massless spring and allowed to hang in the Earth's gravitational field. The spr
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The derived frequency equals 2.63 Hz. Explanation: For an object weighing 8.0 kg with a spring stretching 3.6 cm, calculations involving the spring constant and oscillation frequency lead to this specific oscillation rate.
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Referencing the diagram below, we can deduce from the geometry that x = 2.5 - 0.55 = 1.95 m, leading to cos θ = 1.95/2.5 = 0.78. Therefore, θ = cos⁻¹ 0.78 = 38.74°. According to the free body diagram, the tension in the chain measures 450 N. Here, F denotes the centripetal force and W signifies Dee's weight. The tension's components are as follows: Horizontal component = 450 sin(38.74°) = 281.6 N, directed to the left, and Vertical component = 450 cos(38.74°) = 351.0 N, directed upward. Answers: Horizontal: 281.6, directed left. Vertical: 351.0 N, directed upward.
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Response:

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