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3 years ago

a ball is thrown straight upward and then returns to the earth. does the acceleration change during this motion

1 answer:

Crank3 years ago

5 0

From the time the ball leaves your hand until it hits the ground, the only force acting on it is gravity, and its acceleration is 9.8 m/s^2 downward ... the acceleration of gravity. It doesn't change. By the way, this statement applies whether you throw it or just drop it, and no matter which direction you decide to throw it.

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A bar on a hinge starts from rest and rotates with an angular acceleration α = 12 + 5t, where α is in rad/s2 and t is in seconds

Rama09 [41]

Answer:

Ф = 239.73 rad

Explanation:

α = 12 + 15×t

W = ∫α×dt

= ∫(12 + 5×t)×dt

= 12×t + 2.5×t^2

then:

Ф = ∫W×dt

= ∫(12×t + 2.5×t^2)dt

= 6×t^2 + 5/6×t^3

therefore the angle at t = 4.88s is:

Ф = 6×(4.88)^2 + 5/6×(4.88)^3

= 239.73 rad

5 0

3 years ago

A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend.

MatroZZZ [7]

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2 ( negative because it is pulling the player downwards)

we also know that

s=76 cm= 0.76 m ( maximum s)

using kinetic equation

$v^2=u^2+2as$

where v is final velocity which is zero at max height and u is it initial

hence

$u^2=-2(-9.81)*0.76$

$u=3.8615 m/s\\$

now we can find time in the 15 cm ascent

$s=ut+0.5at^2$

$0.15=3.861*t+0.5*9.81t^2\\$

using quadratic formula

$t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}$

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

$v^2=u^2+2as$

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

$v^2=0^2 +2*(9.81)*(0.76-0.15)$

$v=3.4595$

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

$t=\frac{3.861-3.4595}{9.81}$

t=0.0409

8 0

3 years ago

A chunk of paraffin (wax) has a mass of 50.4 grams and a volume of 57.9 cm3. What is the density of

ExtremeBDS [4]

$Formula\ for\ density:\\\\
p=\frac{m}{V}\\p-density,\\m-mass,\\V-volume\\\\
Data:\\
m=50,4grams\\
V=57,9cm^3\\\\
p=\frac{50,4g}{57,9cm^3}=0,87\frac{g}{cm^3}\\\\Density\ of\ paraffin\ is\ equal\ to\ 0,87\frac{g}{cm^3}.
$

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4 years ago

If a topographic map included a 6,000 ft. mountain next to an area of low hills, which would best describe the contour lines on

VMariaS [17]

The answer to the given question above would be option B. If a topographic map included a 6,000 ft. mountain next to an area of low hills, the statement that best describe the contour lines on the map is this: The contour lines around the mountain would be very close together. Hope this helps.

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Tems11 [23]

Well, the thing is: we don't really know, as we don't even know how many species there are on earth.

If we take a look at the estimates of World Wide Fund for Nature, an organization that works toward combating species extinction, their estimates vary from 200 to 100 000 - but a probable number is 20 000 (d).

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3 years ago

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