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3 years ago

Question 6 (14 points)

Physics

1 answer:

leva [86]3 years ago

3 0

Answer:

$a=114\ m/s^2$

Explanation:

Given that,

Mass of an object, m = 3.68 kg

It is whirled in a horizontal circle of radius 881 cm or 8.81 m

The object completes 10.8 revolutions every 18.8 s.

We need to find the centripetal acceleration of the object. It is given by the formula as follows :

$a=\omega^2 r$ ...(1)

Where $\omega$ is angular velocity

10.8 revolutions = 67.85 rad

$\omega=\dfrac{67.85\ rad}{18.8\ s}\\\\=3.6\ rad/s$

Put values in equation (1)

$a=3.6^2\times 8.81\\\\a=114\ m/s^2$

So, the centripetal acceleration of the object is $114\ m/s^2$ .

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A 20-cm long spring is attached to the wall. When pulled horizontally with a force of 100N, the spring stretches to a length of

suter [353]

Answer:

Explanation:

Part 0

All the spring moves is 2 cm

x = 2 cm * [1 m / 100 cm ]

x = 0.020 meters

F = k*d

100N = k * 0.02 m

100 N / 0.02 = k

5000 N / m

Part A

The spring feels a force of 100 N - - 100N = 200 N because each person is pulling in the opposite direction.

F = k * x

200N = 5000 N/m * d

200 / 5000 = d

d = 0.04 meters.

Part B

10.2 kg must be converted to a force as experienced here on earth.

F = m * g

g = 9.81

m = 10.2

F = 10.2 * 9.81

F = 100.06 N

F = k * d

100.06 = 5000 * d

d = 100.06 / 5000

d = 0.02 meters.

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3 years ago

Chemistry The iron atom (Fe) has 26 protons, 30 neutrons, and 26 electrons. The diameter of the atom is approximately 1.0 × 10 −

Alisiya [41]

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3 years ago

A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be

MrRissso [65]

The kinetic energy halfway the hill is $2.86\cdot 10^5 J$

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

$U_A +K_A = U_B + K_B$

where

$U_A=mgh_A$ is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

$g=9.8 m/s^2$ (acceleration of gravity)

$h_A = 30 m$ (height at point A)

$K_A=\frac{1}{2}mv_A^2$ is the initial kinetic energy, at point A , with

$v_A=17 m/s$ (velocity at point A)

$U_B=mgh_B$ is the final potential energy, at point B, where

$h_B = 15 m$ (height at point B)

$K_B=\frac{1}{2}mv_B^2$ is the final kinetic energy, at point B, where

$v_B$ is the velocity at point B

Here we are interested in finding $K_B$ , so by re-arranging the equation and substituting we find:

$K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J$

Learn more about kinetic energy:

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8 0

3 years ago

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 6.0 rev/s in 11.0 s. At th

sleet_krkn [62]

Answer

given,

ω₁ = 0 rev/s

ω₂ = 6 rev/s

t = 11 s

Using equation of rotational motion

The angular acceleration is

ωf - ωi = α t

11 α = 6 - 0

= 0.545 rev/s²

The angular displacement

θ₁= ωi t + (1/2) α t²

θ₁= 0 + (1/2) (0.545)(11)^2

θ₁= 33 rev

case 2

ω₁ = 6 rev/s

ω₂ = 0 rev/s

t = 14 s

Using equation of rotational motion

The angular acceleration is

ωf - ωi = α t

14 α = 0 - 6

= - 0.428 rev/s²

The angular displacement

θ₂= ωi t + (1/2) α t²

θ₂= 6 x 14 + (1/2) (-0.428)(14)^2

θ₂= 42 rev

total revolution in 25 s is equal to

θ = θ₁ + θ₂

θ = 33 + 42

θ = 75 rev

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3 years ago

if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?

madreJ [45]

Answer:

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

$a_{c} =\frac{(velocity)^{2} }{radius}$

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

$a_{c} = ?$

Solution:

We have,

$a_{c} =\frac{(velocity)^{2} }{radius}$

Substituting given value in it we get

$a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2$

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

7 0

3 years ago