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Dominik [7]
3 years ago
12

Three objects of uniform density—a solid sphere, a solid cylinder, and a hollow cylinder— are placed at the top of an incline (f

ig. cq10.13). they are all released from rest at the same elevation and roll without slipping. (a) which object reaches the bottom first

Physics
1 answer:
TiliK225 [7]3 years ago
5 0
First you must make a free body diagram of each body to find the acceleration of each one. After you find the general formula of acceleration, you must replace the values of K for each body. (Assuming they all have the same mass). Finally, by replacing K, the body that gets the most acceleration will be the first to reach the bottom.I attached the solution.In this case the solid sphere reaches the bottom first.

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A basketball player can leap upward 0.43 m. how long does he remain in the air? use an acceleration due to gravity of 9.80 m/s2
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From the equations of linear motion,
v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,
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                = 2×9.81×0.43
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3 years ago
What is the current when the resistance is 5 ohms and the voltage is 10 volts?
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Ohm's Law states V = IR
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The mass of a hot-air balloon and its cargo (not including the air inside) is 170 kg. The air outside is at 10.0°C and 101 kPa.
scoundrel [369]

Answer:

108.37°C

Explanation:

P₁ = Initial pressure = 101 kPa

V₁ = Initial volume = 530 m³

T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K

P₂ = Final pressure = 101 kPa (because it is open to atmosphere)

V₂ = Final volume = 530 m³

P₁V₁ = n₁RT₁

⇒101×530 = n₁RT₁

⇒53530 J = n₁RT₁

P₂V₂ = n₂RT₂

⇒53530 J = n₂RT₂

\frac{m_1}{m_2}=\frac{\rho V_1}{\rho V_1-170}\\\Rightarrow \frac{m_1}{m_2}=\frac{1.244\times 530}{1.244\times 530-170}=1.347\\\Rightarrow \frac{m_1}{m_2}=1.347\\\Rightarrow \frac{n_1}{n_2}=1.347

Dividing the first two equations we get

1=\frac{n_1}{n_2}\frac{T_1}{T_2}\\\Rightarrow 1=1.347\frac{283.15}{T_2}\\\Rightarrow T_2=1.347\times 283.15= 381.52\ K

∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C

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3 years ago
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