All categories

3 years ago

Assume the height of the roller coaster (see fig) is y = 40 m (take the reference point, y = 0, as the bottom of the

Physics

1 answer:

PtichkaEL [24]3 years ago

7 0

H⁣⁣⁣⁣ere's l⁣⁣⁣ink t⁣⁣⁣o t⁣⁣⁣he a⁣⁣⁣nswer:

bit. $^{}$ ly/3a8Nt8n

You might be interested in

A laborer pushes a box to distance of 20m. if the force required to push the box is 25N what will be the quantity of work done

Marrrta [24]

Answer:

500 J

Explanation:

In the case of a rectilinear movement, the work is calculated as the product of Force (N) * movement (m). In your case, unless the angle between the force vector and the displacement vector is different from 0, the work is:

25 N * 20 m = 500 J

7 0

2 years ago

A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translat

Talja [164]

Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

$m g =\dfrac{mv^2}{r}$

$9.81 =\dfrac{v^2}{0.675}$

v² = 6.622

v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

where, $I =\dfrac{2}{5}mr^2$

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

$0.7 u^2 + g H = 0.7 v^2 + g(2R)$

$0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)$

0.7 u² = 7.85092

u² = 11.2156

u = 3.35 m/s

the initial speed is 3.35 m/s

8 0

2 years ago

A young man exerted a force of 9,000 N on a stalled car but was unable to move it. How much work was done?

olasank [31]

W=F*s
W=9000*0
W=0
W:Work
F:Force
s:Distance

5 0

2 years ago

An object is acted upon by a force of 22 newtons to the right and a force of 13 newtons to the left. What is the magnitude and d

ale4655 [162]

<span>An object is acted upon by a force of 22 newtons to the right and a force of 13 newtons to the left.

(1) 22 N to the right
(2) 13 N to the left.

magnitude = 22 - 13
magnitude = 9

Direction would be to the right.

So magnitude is 9N direction to the right.</span>

8 0

3 years ago

I need help with this question

Juli2301 [7.4K]

Answer:

Centripetal force is the force that keeps the yoyo going in a circle, if the string breaks, the yoyo would would fly off in a direction that is different to the point on the circle.

8 0

3 years ago