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Readme [11.4K]
3 years ago
5

Basidiomycete fungi ballistically eject millions of spores into the air by releasing the surface tension energy of a water dropl

et condensing on the spore. The spores are ejected with typical speeds of 1.09 m/s, allowing them to clear the "boundary layer" of still air near the ground to be carried away and dispersed by winds.
(a) If a given spore is accelerated from rest to 1.09 m/s in 7.85 µs, what is the magnitude of the constant acceleration of the spore (in m/s^2) while being ejected?
Physics
1 answer:
GalinKa [24]3 years ago
6 0

Answer:

1.4 x 10^5 m/s^2.

Explanation:

Given the following:

vi = 0 m/s.

vf = 1.09 m/s.

t = 7.85 µs.

Using,

vf = vi + a*t

1.09 = a*7.8 x 10^-6

a = 139743.6

= 1.4 x 10^5 m/s^2.

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A shopper pushes a 5.32 kg grocery cart
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Answer:

\text { acceleration of the cart is } 10.94 \mathrm{m} / \mathrm{s}^{2}

Explanation:

According to “Newton's second law”

“Force” is “mass” times “acceleration”, or F = m× a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass

Force = mass × acceleration

\text { Acceleration }=\frac{\text { force }}{\text { mass }}

Given that,

Mass = 5.32 kg

\text { Force }=12.7 \mathrm{N} \text { forces at }-28.7^{\circ}

x=-28.7^{\circ}

F = 12.7N

Normal force = mg + F sinx,  

“m” being the object's "mass",  

“g” being the "acceleration of gravity",

“x” being the "angle of the cart"

\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^ 2 \text { on Earth })

To find normal force substitute the values in the formula,

Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)

Normal force = 52.136 + 12.7 × 0.480

Normal force = 52.136 + 6.096

Normal force = 58.232 N

<u>Acceleration of the cart</u>:

\text { Acceleration }=\frac{\text {Normal force}}{\text { mass }}

\text { Acceleration }=\frac{58.232}{5.32}

\text { Acceleration }=10.94 \mathrm{m} / \mathrm{s}^{2}

\text { Therefore, "acceleration of the cart" is } 10.94 \mathrm{m} / \mathrm{s}^{2}

7 0
2 years ago
Read 2 more answers
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