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nata0808 [166]
3 years ago
15

In Newton’s second law, if the net force acting on object doubles. The object’s Will also double

Physics
1 answer:
Anna71 [15]3 years ago
3 0

Answer: Explanation:

If the net force on an object is doubled, its acceleration will double If the mass of an object is doubled, the acceleration will be halved .

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An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic
malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

Where,

\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

x = 14.397*(2\pi*\frac{0.25}{2})

x = 11.3m

5 0
3 years ago
What happens to the coefficient of friction when the weight is increased? Why is this?
Crazy boy [7]

Answer:

Usually the coefficient of friction remains unchanged

Explanation:

The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.

Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.

6 0
3 years ago
Read 2 more answers
What is the least force you could exert on the refrigerator to move it?
Ulleksa [173]
 <span>If the refrigerator weights 1365 and you are not exerting any vertical force on it, then the normal force is also 1365N. so Fn=1365 

Fsf = Static frictional force = (coefficient of static friction) * (Normal force) 

So the least for you could exert to move it is equal to the Fsf. 
Fsf = (0.49)(1365N)</span><span>
</span>
8 0
3 years ago
B. The silica cylinder of a radiant wall heater is 0.6 m long
SIZIF [17.4K]

So,  If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.

<h3>Power radiated by the radiant wall heater</h3>

The power radiated by the radiant wall heater is given by P = εσAT⁴ where

  • ε = emissivity = 1 (since we are not given),
  • σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
  • A = surface area of cylindrical wall heater = 2πrh where
  • r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
  • h = length of heater = 0.6 m, and
  • T = temperature of heater

Since P = εσAT⁴

P = εσ(2πrh)T⁴

Making T subject of the formula, we have

<h3>Temperature of heater</h3>

T = ⁴√[P/εσ(2πrh)]

Since P = 1.5 kW = 1.5 × 10³ W

Substituting the values of the variables into the equation, we have

T = ⁴√[P/εσ(2πrh)]

T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]

T = ⁴√[1.5 × 10³ W/(43.2π  × 10⁻¹¹ W/K⁴)]

T = ⁴√[1.5 × 10³ W/135.72  × 10⁻¹¹ W/K⁴)]

T = ⁴√[0.01105 × 10¹⁴ K⁴)]

T = ⁴√[1.105 × 10¹² K⁴)]

T = 1.0253 × 10³ K

T = 1025.3 K

So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

Learn more about temperature of radiant wall heater here:

brainly.com/question/14548124

6 0
2 years ago
How are domains arranged in materials that are magnetic and in ones that are not
Valentin [98]

Answer:

In most materials, atoms are arranged in such a way that the magnetic orientation of one electron cancels out the orientation of another

8 0
3 years ago
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