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3 years ago

15

A student makes a simple pendulum by attaching a mass to the free end of a 1.50-meter length of string

1 answer:

balu736 [363]3 years ago

7 0

-- First, it can never swing to a higher elevation off the floor
than where it was when she let it go. The higher it is off the
floor, the more potential energy it has, and that's all the energy
she gave it when she lifted it to the height of her chin.

-- Second, it can't even return to THAT height, because during
its swing out and back, it's losing energy by plowing through air.
So each swing is slightly narrower, and ends slightly lower, than
the one before it.

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4 years ago

Frank’s automobile engine runs at 100∘C. One day, when the outside temperature is 15∘C, he turns off the ignition and notes that

lapo4ka [179]

Answer:

the engine cool to 40 $^{o}C$ at 14.07 minutes

Explanation:

Given information

T(5) = 70 $^{o}C$

$T_{0}$ = 100 $^{o}C$

C = 15 $^{o}C$

Newton's law of cooling :

T(t) = C + ( $T_{0}$ - C) $e^{-kt}$

where

T(t) = temperature at any given time

C = surrounding temperature

$T_{0}$ = initial temperature of heated object

k = cooling constant

to find the the time when the engine will be cooled down to 40 $^{o}C$ , we first need to find the cooling constant, k

when t = 5, T(5) = 70 $^{o}C$

so,

T(t) = C + ( $T_{0}$ - C) $e^{-kt}$

T(5) = 15 + (100 - 15) $e^{-5k}$

70 = 15 + (85) $e^{-5k}$

$e^{-5k}$ = (70 - 15) / 85

-5k = ln (55/85)

k = - ln (55/85) / 5

k = 0.087

thus, we have the eqaution

T(t) = 15 + (85) $e^{-0.087t}$

now we can determine the time when T(t) = 40 $^{o}C$

40 = 15 + (85) $e^{-0.087t}$

$e^{-0.087t}$ = (40-15)/85

-0.0087t = ln (25/85)

t = - ln (25/85)/0.087

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8 0

3 years ago

A 2,200 kg railway freight car coasts at 3.6 m/s underneath a grain terminal, which dumps grain directly down into the freight c

suter [353]

Answer:

$m_2=2554.83\ kg$

Explanation:

Given that,

Mass of the car, $m_1=2200\ kg$

Initial speed of the car, $u=3.6\ m/s$

Final speed of the car, $v=3.1\ m/s$

To find,

The maximum mass of grain that it can accept.

Solution,

We know that according to the law of conservation of momentum, the initial momentum is equal to the final momentum.

$m_1u=m_2v$

$m_2=\dfrac{m_1u}{v}$

$m_2=\dfrac{2200\times 3.6}{3.1}$

$m_2=2554.83\ kg$

Therefore, the maximum mass of grain that it can accept is 2554.83 kg. Hence, this is the required solution.

7 0

3 years ago