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Novosadov [1.4K]

2 years ago

9

What is the answer to 895cm a Km

1 answer:

victus00 [196]2 years ago

8 0

Answer:

if you're converting then the answer is 0.00895

Explanation:

895 centimetres converted into kilometres= 0.00895

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A child walks 5.0 meters north, then 4.0 meters east, and finally 2.0 meters south. what is the magnitude of the resultant displ

tensa zangetsu [6.8K]

Hi, if you draw a diagram of what is actually happening you will be able to understand what is happening much easier. Also displacement means distance from the starting point not the distance traveled if you didn't know already. so |dw:1318897040544:dw| if you follow this diagram you can see how the child has walked, you can then make a triangle and work the displacement which in the case is the root of the sum of the two other sides. This is just Pythagoras a^2+b^2=c^2 2

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Which organism would be a top consumer in a food chain

Kruka [31]

Answer: A herbivore.

Explanation:

A herbivore would be a top consumer in a food chain because they are primary consumers which leads you to secondary consumers who are carnivores to make sure they don't get to over populated and keep balance.

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3 years ago

A ball on the end of a rope is moving in a vertical circle near the surface of the earth. Point A is at the top of the circle; C

mrs_skeptik [129]

Answer:

Tension in the string will increase

Explanation:

As we know that tension in the string at any angle with the vertical is given as

$T - mgcos\theta = m\omega^2 R$

now we have

$T = mgcos\theta + m\omega^2 R$

also we know that

angular speed of the stone is directly depending on the time period of the motion

so it is given as

$\omega = \frac{2\pi}{T}$

since the frequency of the revolution is increased from n = 1 rev/s to 2 rev/s

so the angular speed would be doubled

So here we can say that

tension in the string will increase when we will increase the frequency of revolution.

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3 years ago

A 0.144-kg baseball is moving toward home plate with a speed of 43 m/s when

algol [13]

I would say 648858. bc yes

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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