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CaHeK987 [17]
3 years ago
13

A tiger leaps with an initial velocity of 35.0 km/hr at an angle of 13.0ᶿ with respect to the horizontal. What are the component

s of the tiger’s velocity?
Physics
1 answer:
Lorico [155]3 years ago
3 0

Answer:

Vx = 35 x cos(13deg)

Vy = 35 x sin(13deg) - gt  

(g is acceleration due to gravity =~9.8 meter/second^2, t is time in second)

Explanation:

The tiger leaps up, then x and y component of its velocity are:

Vx = Vo x cos(alpha)

Vy = Vo x sin(alpha) - gt

(Vo is tiger's initial velocity, alpha is angle between its leaping direction and horizontal plane)

Hope this helps!

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cestrela7 [59]

Answer:

T_{2}=278.80 K

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

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\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}

Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

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6 0
3 years ago
A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),
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Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

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W = F.d

W = F d Cosθ

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W = Work Done

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d = displacement

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Since, force P is parallel to the motion of the box. Therefore, θ = 0°

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W = Pd

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<u></u>

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Hence,

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