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Furkat [3]
2 years ago
6

A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an

d tangential components of acceleration of the discus when the angular velocity is 10 rad/s. (80 m/s2, 40 m/s2).
Physics
1 answer:
anyanavicka [17]2 years ago
7 0

Answer:

The value of tangential acceleration \alpha_{t} =  40 \frac{m}{s^{2} }

The value of radial acceleration \alpha_{r} = 80 \frac{m}{s^{2} }

Explanation:

Angular acceleration = 50 \frac{rad}{s^{2} }

Radius of the disk = 0.8 m

Angular velocity = 10 \frac{rad}{s}

We know that tangential acceleration is given by the formula \alpha_{t} = r \alpha

Where r =  radius of the disk

\alpha = angular acceleration

⇒ \alpha_{t} = 0.8 × 50

⇒ \alpha_{t} = 40 \frac{m}{s^{2} }

This is the value of tangential acceleration.

Radial acceleration is given by

\alpha_{r} = \frac{V^{2} }{r}

Where V = velocity of the disk = r \omega

⇒ V = 0.8 × 10

⇒ V = 8 \frac{m}{s}

Radial acceleration

\alpha_{r} = \frac{8^{2} }{0.8}

\alpha_{r} = 80 \frac{m}{s^{2} }

This is the value of radial acceleration.

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Answer:

m_l=550\ kg is the mass of librarian.

Explanation:

Given:

  • mass of the system, m_s=3.3\times 10^{3}\ kg
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N<u>ow using the principle of elastic collision:</u>

Net momentum of the system is zero.

m_l\times v_l=(3300-m_l)\times v_c

m_l\times 2.5=(3300-m_l)\times 0.5

m_l=550\ kg is the mass of librarian.

6 0
2 years ago
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Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5
Bogdan [553]
The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

     E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

   E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.
8 0
2 years ago
A hoop is rolling (without slipping) on a horizontal surface so it has two types of kinetic energy: translational kinetic energy
LenaWriter [7]

Answer:

\dfrac{T}{K}=1

Explanation:

The translational kinetic energy of the hoop is given by :

K=\dfrac{1}{2}Mv^2..................(1)

M is the mass of the hoop

v is the velocity of the hoop

The rotational kinetic energy of the hoop is given by :

T=\dfrac{1}{2}I\omega^2

Since, I=MR^2

\omega=\dfrac{v}{R}

T=\dfrac{1}{2}\times MR^2\times (\dfrac{v}{R})^2..............(2)

From equation (1) and (2) :

\dfrac{T}{K}=1

Therefore, the ratio of the translational kinetic energy to the rotational kinetic energy is 1.

8 0
2 years ago
(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45
Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m

The height the tomato would fall is 450+577.2 = 1027.2 m

6 0
3 years ago
A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc
Dafna11 [192]

Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

First we find the acceleration of the centrifuge using,

a = (v-u)/t......................... Equation 1

Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s,  u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

Where α = angular acceleration, r = radius of the centrifuge

Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

therefore,

Ф = (u/r+v/r).t/2

where u = 0 m/s (at rest),  = 150 m/s, r = 0.08 m, t = 100 s

Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

If,

1 rad = 0.159155 rev,

Ф = (93750×0.159155) rev

Ф = 14920.78 rev

6 0
2 years ago
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