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sp2606 [1]
3 years ago
7

How can you say gravitational force is a field force

Physics
1 answer:
Lana71 [14]3 years ago
6 0

Answer:

because gravity pulled us in the land if there is no gravitational force there will not be field force too

Explanation:

hope it's will help you

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A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

  • If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
  • τ = F*r (2)
  • For a solid uniform disk, the rotational inertia regarding an axle passing through its center  is just I = m*r²/2 (3).
  • Replacing (2) and (3) in (1), we can solve for α, as follows:

       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

        \omega_{f}^{2}  - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)

  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
  • Replacing this value and (7) in (8), we get:

       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

b)    

  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

       a_{t} = \alpha * r (9)

  • where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
  • Replacing this value and (4), in (9), we get:

       a_{t}  = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)

  • Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
  • The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

       a_{c} = \omega^{2} * r  (11)

  • Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
  • Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
  • Replacing this value and (7) in (11) we get:

       a_{c} = \omega^{2} * r   = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)

  • The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
  • Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

6 0
2 years ago
A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
2 years ago
Of the following systems, which contains the most heat?
densk [106]

Answer:d

Explanation:

Given systems are state of matter and do not contain any heat instead Heat is required to change Phase or raise the temperature of the particular system.

For example 600 kg of ice at 0^{\circ}C

Heat Required to convert it to water at 0^{\circ}C is

Q=m\times L

Where L=latent heat of Fusion =334 J/gm

Q=600\times 334\times 1000

Q=200.4 MJ

8 0
3 years ago
What replaces a cold current that sinks to the ocean floor?
FrozenT [24]
Well simple the warm water then replaces the cold current that sinks to the ocean floor.
5 0
3 years ago
Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 45.0 mph . Beth leaves Los A
WARRIOR [948]

Answer:

a.Beth

b.2232 s

Explanation:

We are given that

Distance,d=400 mi

Speed of Alan,v=45 mph

Speed of Beth,v'=55 mph

a.Time =\frac{distance}{speed}

Using the formula

Time taken by Alan=\frac{400}{45}=8.89 hr

Time taken by Beth=\frac{400}{55}=7.27hr

Alan will reach San Francisco at 4:53 PM

Beth will reach San Francisco at 4:16 PM

Beth will reach before Alan.

b.Difference between time=8.89-7.27=1.62 hr

t=1.62 hr

1.62-1=0.62 hr

0.62 hr=0.62\times 60\times 60=2232 s

Hence, Beth has to wait 2232 s for Alan to arrive .

6 0
3 years ago
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