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gayaneshka [121]

3 years ago

14

A skier leaves the horizontal end of a ramp with a velocity of 31.0 m/s and lands 156.3 m from the base of a ramp how high is th

e end of the ramp from the ground ?

1 answer:

BartSMP [9]3 years ago

3 0

<u>Answer:</u>

The height of ramp = 124.694 m

<u>Explanation:</u>

Using second equation of motion,

$s = ut + \frac{1}{2}at^2$

From the question,

u = 31 m/s; s = 156.3 m, a=0

substituting values

$156.3 = 31\times t + 0$

t = $\frac{156.3}{31 }$

= 5.042 s

Similary, for the case of landing

t = 5.042 s; initial velocity, u =0

acceleration = acceleration due to gravity, g = 9.81 $m/s^2$

Substituting in $h = ut + \frac{1}{2}gt^2$

$h = 0 + \frac{1}{2} \times 9.81 \times (5.042)^2$

h = 124.694 m

So height of ramp = 124.694 m

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The given function represents the position of a particle traveling along a horizontal line. s(t) = 2t3 − 3t2 − 12t + 6 for t ≥ 0

avanturin [10]

Answer:

(a) $v(t) = 6t^2 - 6t - 12$ , $a(t) = 12t - 6$

(b) When $0 \leq t < 0.5$ , object is slowing down, when $t > 0.5$ object is speeding up.

Explanation:

(a) To get the velocity function, we need to take the derivative of the position function.

$v(t) = \frac{ds(t)}{dt} = (2t^{3})^{'} - (3t^{2})^{'} - (12t)^{'} + 6^{'} = 6t^{2} - 6t - 12$

To get the acceleration function, we need to take the derivative of the velocity function.

$a(t) = \frac{dv(t)}{dt} = (6t^{2})^{'} - (6t)^{'} - (12)^{'} = 12t - 6$

(b) The object is slowing down when velocity is decreasing by time (decelerating) hence a < 0

$12t - 6 < 0 \\12t < 6 \\t < 0.5$

On the other hand, object is speeding up when a > 0

$12t - 6 > 0 \\12t > 6 \\t > 0.5$

Therefore, when $0 \leq t < 0.5$ , object is slowing down, when $t > 0.5$ object is speeding up.

6 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

ONLY TWO ANSWER CHOICES!!

Otrada [13]

A: Buoyant force is equal to the weight

3 0

3 years ago

Please<br>help<br>me<br>with this problem

uranmaximum [27]

Answer:

speeding up

Explanation:

because its speeding up, theres going to be more newtons in the back

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2 years ago

Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit

cestrela7 [59]

Answer:

$v=1667.9km/h$

$a_{cp}=436.6km/h^2$

Explanation:

The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth $C=2\pi R$ , where $R=6371km$ is the radius of the Earth.

We have then:

$v=\frac{C}{t}=\frac{2\pi R}{t}=\frac{2\pi (6371km)}{(24h)}=1667.9km/h$

And then we use the centripetal acceleration formula:

$a_{cp}=\frac{v^2}{R}=\frac{(1667.9km/h)^2}{(6371km)}=436.6km/h^2$

6 0

3 years ago