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Eva8 [605]
2 years ago
9

How can you determine the amount of work done on an object? (joules, work, power...etc.)

Physics
1 answer:
Novay_Z [31]2 years ago
7 0
Work done is equal to force by distance; so you take the force exerted, in newtons, and multiply that by the direction it's moved (from the starting point in a line, not along the path it's taken.)
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Find the angular momentum of a hollow spinning sphere when its angular speed is 20rad/s. The rotational inertia of the sphere is
zysi [14]

Answer:

L= 0.4 kgm²/s

Explanation:

The angular momentum of a hallow spinning sphere is

L = Iω

L = (0.02kgm²) × (20rad/s)

L= 0.4kgm²/s

5 0
2 years ago
A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has
irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration (a) consist in two components, <em>radial</em> (a_{r}) and <em>tangential</em> (a_{t}), in meters per square second:

a_{r} = \omega^{2}\cdot r (1)

a_{t} = \alpha \cdot r (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

a = \sqrt{a_{r}^{2}+a_{t}^{2}} (3)

Where:

r - Radius of the wheel, in meters.

\omega - Angular speed, in radians per second.

\alpha - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

\omega = \omega_{o}+ \alpha\cdot t (4)

Where:

\omega_{o} - Initial angular speed, in radians per second.

t - Time, in seconds.

If we know that r = 0.1\,m, \alpha = 2\,\frac{rad}{s^{2}}, \omega_{o} = 0\,\frac{rad}{s} and t = 0.60\,s, then the total linear acceleration is:

\omega = \omega_{o}+ \alpha\cdot t

\omega = 1.2\,\frac{rad}{s}

a_{r} = \omega^{2}\cdot r

a_{r} = 0.144\,\frac{m}{s^{2}}

a_{t} = \alpha \cdot r

a_{t} = 0.2\,\frac{m}{s^{2}}

a = \sqrt{a_{r}^{2}+a_{t}^{2}}

a \approx 0.246\,\frac{m}{s^{2}}

The total linear acceleration is approximately 0.246 meters per square second.

3 0
3 years ago
The angular speed of digital video discs (DVDs) varies with whether the inner or outer part of the disc is being read. (CDs func
Kitty [74]

Answer:

α = 0.0135 rad/s²

Explanation:

given,

t = 133 min = 133 x 60 = 7980 s

angular speed varies from 570 rpm to 1600 rpm

now,

570 rpm = 570 \times \dfrac{2\pi}{60}

              = 59.69 rad/s

1600 rpm =  = 570 \times \dfrac{2\pi}{60}

              = 167.6 rad/s

using equation of rotational motion

ωf = ωi + αt    

167.6 = 59.7 + α x 7980

α  x 7980 = 107.9

α = 0.0135 rad/s²

8 0
3 years ago
High-speed stroboscopic photographs show that the head of a 210-g golf club is traveling at 56 m/s just before it strikes a 46-g
tamaranim1 [39]

Explanation:

It is given that,

Mass of golf club, m₁ = 210 g = 0.21 kg

Initial velocity of golf club, u₁ = 56 m/s

Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg

After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.

Initial momentum of golf ball, p_i=m_1u_1=0.21\ kg\times 56\ m/s=11.76\ kg-m/s

After the collision, final momentum p_f=0.21\ kg\times 42\ m/s+0.046v

Using the conservation of momentum as :

p_i=p_f

11.76\ kg-m/s=0.21\ kg\times 42\ m/s+0.046v

v = 63.91 m/s

So, the speed of the  golf ball just after impact is 63.91 m/s. Hence, this is the required solution.

3 0
3 years ago
Find the Input force, if the Mechanical Advantage of the simple machine used is 5 and Output force is 50 N.
galina1969 [7]

Answer:

6N

Explanation:

6 0
3 years ago
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