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3 years ago

MATHPHYS HELP

Physics

1 answer:

rusak2 [61]3 years ago

3 0

Answer:

16.4287

Explanation:

The force and displacement are related by Hooke's law:

F = kΔx

The period of oscillation of a spring/mass system is:

T = 2π√(m/k)

First, find the value of k:

F = kΔx

78 N = k (98 m)

k = 0.796 N/m

Next, find the mass of the unknown weight.

F = kΔx

m (9.8 m/s²) = (0.796 N/m) (67 m)

m = 5.44 kg

Finally, find the period.

T = 2π√(m/k)

T = 2π√(5.44 kg / 0.796 N/m)

T = 16.4287 s

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A baseball is moving at a speed of 2.2m/s when it strikes the catchers glove. The paddding of the glove is compressed by 24mm be

Andre45 [30]

Average acceleration of the baseball: $-101 m/s^2$

Explanation:

Since the motion of the baseball is a uniformly accelerated motion, we can use the following suvat equation:

$v^2-u^2=2as$

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the displacement of the object

For the baseball in this problem, we have:

u = 2.2 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

s = 24 mm = 0.024 m is the displacement of the ball while decelerating

Therefore, we can solve for a to find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{0-2.2^2}{2(0.024)}=-101 m/s^2$

where the negative sign means the baseball is slowing down.

Learn more about acceleration:

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3 years ago

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m_a_m_a [10]

Answer of your question is in this photo

8 0

2 years ago

Number 10 and an explaination would be fabulous. thanks!

Sati [7]

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3 0

3 years ago

Read 2 more answers

Be sure to answer all parts. By what factor does the fraction of collisions with energy equal to or greater than activation ener

qwelly [4]

Answer:

Factor = 8.77

Explanation:

Fraction of collision with energy greater than or equal to the activation energy can be given by the formula:

$f = \exp(\frac{-E_{a} }{RT} )$

$E_{a} = Activation Energy\\E_{a} = 100 kJ /mol\\E_{a} = 10^{5} J /mol\\$

R = 8.314 J/mol.K

When Temperature = 34⁰C = 34 + 273 = 307 K

$f_{1} = \exp(\frac{-10^{5} }{8.314 * 307} )\\f_{1} = \exp(\frac{-10^{5} }{2552.398} )\\f_{1} = \exp(-39.18)\\f_{1} = 964.59 * 10^{-20}$

When Temperature = 52⁰C = 52 + 273 = 325 K

$f_{2} = \exp(\frac{-10^{5} }{8.314 * 325} )\\f_{2}= \exp(\frac{-10^{5} }{2702.05} )\\f_{2}= \exp(-37.009)\\f_{2} = 8456.6 * 10^{-20}$

$\frac{f_{2}}{f_{1}}= \frac{8456.6 * 10^{-20} }{964.59 * 10^{-20} }$

Factor = 8.77

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3 years ago

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kirill115 [55]

To solve this problem we need to use the concepts related to fluid density. Since we need to know a height point of two different liquids, their pressures must be equal, so

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