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Troyanec [42]
3 years ago
11

A 120-kg roller coaster cart is being tested on a new track, and a crash-test dummy is loaded into it. The roller coaster starts

from rest at point A which is 25 m above point B. In your own words, what physics ideas would you be able to use to predict how fast the cart is moving at point B? What calculations would you use, what things would you be measuring, and what physics principles or laws would you use to make this calculation
Physics
1 answer:
avanturin [10]3 years ago
5 0

Answer:

vb = 22.13 m/s

So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.

Explanation:

In order to find the speed of roller coaster at Point B, we will use the law of conservation of Energy. In this situation, the law of conservation of energy states that:

K.E at A + P.E at A = K.E at B + P.E at B

(1/2)mvₐ² + mghₐ = (1/2)m(vb)² + mg(hb)

(1/2)vₙ² + ghₐ = (1/2)(vb)² + g(hb)

where,

vₙ = velocity of roller coaster at point a = 0 m/s

hₙ = height of roller coaster at point a = 25 m

g = 9.8 m/s²

vb = velocity of roller coaster at point B = ?

hb = Height of Point B = 0 m (since, point is the reference point)

Therefore,

(1/2)(0 m/s)² + (9.8 m/s²)(25 m) = (1/2)(vb)² + (9.8 m/s²)(0 m)

245 m²/s² * 2 = vb²

vb = √(490 m²/s²)

<u>vb = 22.13 m/s</u>

<u>So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.</u>

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Calculate 8 ∙ 10^-4 divided by 2 ∙ 10^2.
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3 years ago
An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4.00 s.
Veseljchak [2.6K]

Answer:

a) -1.25 rev/s² and 23.3 rev

b)  2.67s

Explanation:

a) ωФ_o_z = (500 rev/min)(1min/ 60s) => 8.333 rev/s

ωФ_Z= (200 rev/min)(1min/ 60s) => 3.333rev/s

time 't'= 4 s

angular acceleration 'αФ_Z'=?

constant angular acceleration equation is given by,

ωФ_Z= ωФ_o_z + αФ_Zt

αФ_Z= (ωФ_Z - ωФ_o_z )/t => (3.333-8.333)/4

αФ_Z= -1.25 rev/s²

θ-θФ_o = ωФ_o_z t + 1/2αФ_Zt²

      =(8.333)(4) + 1/2 (-1.25)(4)²

      =23.3 rev

b) ωФ_Z=0   (comes to rest)

ωФ_o_z = 3.333 rev/s

αФ_Z= -1.25 rev/s²

ωФ_Z= ωФ_o_z + αФ_Zt

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3 0
3 years ago
A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails
harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

y=h+ut+\frac{1}{2}gt^2

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

v^2-u^2 = 2ad

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

a=\frac{v-u}{t}

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

8 0
3 years ago
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