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nika2105 [10]
3 years ago
6

Ms. Kasper is in a panic. Her cat, Penny, is stuck in a tree and about to jump out. In order to save her cat, Ms. Kasper needs t

o run to the tree, 12 meters away. If it takes her cat, 3 seconds to fall, how fast would Ms. Kasper have to run to save her cat?
Physics
1 answer:
Misha Larkins [42]3 years ago
3 0

Answer:

Speed greater than 4 m/s

Explanation:

Given that Ms. Kasper is in a panic. Her cat, Penny, is stuck in a tree and about to jump out. In order to save her cat, Ms. Kasper needs to run to the tree, 12 meters away. If it takes her cat, 3 seconds to fall, how fast would Ms. Kasper have to run to save her cat?

The distance = 12 m

Time = 3s

Speed = distance/time

Speed = 12/3

Speed = 4 m/s

Ms Kasper must run at speed more than 4m/s for her to save the cat.

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Answer:

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Explanation:

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\Phi = EA cos \theta

where

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A is the cross-sectional area

\theta is the angle between the direction of the electric field and the normal to A

The flux is maximum when \theta=0^{\circ}, so we are in this situation and therefore cos \theta =1, so we can write

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Here we have:

\Phi = 7.50\cdot 10^5 N/m^2 C is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

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And so, we can find the magnitude of the electric field:

E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C

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Answer:

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Explanation:

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How to find out the heat capacity of a material?​
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\huge\underline{\underline{\boxed{\mathbb {EXPLANATION}}}}

The heat capacity is given by the expression:

\longrightarrow \sf{\triangle Q= m \triangle C  \triangle   T}

\longrightarrow \sf{Q= \: Heat}

\longrightarrow \sf{M= \: Mass}

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\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}

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