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4 years ago

10

What happens when a roller coaster car moves down from the top of a hill?

2 answers:

VARVARA [1.3K]4 years ago

6 0

-- It accelerates.
-- Its speed increases.
-- It gains momentum.
-- It loses altitude.
-- It loses potential energy.
-- It gains kinetic energy.
-- Its wheels make a lot of noise.
-- Everybody screams.

Stella [2.4K]4 years ago

3 0

Potential energy is transformed into kinetic and thermal energy.

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(PLEASE HELP THANK YOU) A car is going 8 meters per second on an access road into a highway and then accelerates at 1.8 meters p

jonny [76]

Answer:

20.96 m/s

Explanation:

Apply the kinematic equation:

Vf=Vi+at

Vi=8m/s

a=1.8m/s^2

t=7.2s

Putting this all in should give you your answer of 12.96m/s

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3 years ago

The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom ris

Andrew [12]

Explanation:

Let us assume that forces acting at point B are as follows.

$\sum F_{x} = 0$

$T + F_{AB} Sin 60$ = 0 ...... (1)

$\sum F_{y}$ = 0

$F_{AB} Cos 60 + W$ = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

$\sigma_{allow} = \frac{T}{A}$

T = $(20 \times 1000) \frac{\pi}{4} \times (0.5)^{2}$

= 3925 kip

From equation (1), $F_{AB}Sin (60^{o})$ = -3925

$F_{AB} \times -0.304 8$ = -3925

$F_{AB}$ = 12877.29 kip

From equation (2), -12877.29 (Cos 60) + W = 0

$-12877.29 kip \times \frac{1}{2} + W$ = 0

W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

5 0

3 years ago

2. (1) A piece of rubber is 50 cm long when a weight of

Harrizon [31]

Answer

By F = -kx {-ve just indicating the sign of the force}

=>35 = k x (85-50) x 10^-2

=>k = 100 N/m

Again by F = -kx

8 0

3 years ago

PLEASE HELP ME ASAP PLEASEEEEE

AleksAgata [21]

I believe its the law of inertia

6 0

3 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago