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Scilla [17]
2 years ago
13

An object moves with a constant speed of 20 m/s on a circular track of radius 100 m. What is the tangential acceleration of the

object?
Physics
1 answer:
MissTica2 years ago
4 0

Answer:4s^-1

Explanation:Tangential Acceleration=v^2/r=(20)²/100=400/100

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An object of mass 6 kg. is resting on a horizontal surface. A horizontal force
son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force (W), measured in joules, is defined by the following expression:

W = F\cdot \Delta x (1)

Where:

F - Force, measured in newtons.

\Delta x - Distance, measured in meters.

If we know that F = 15\,N and \Delta x = 100\,m, then the work done by the force exerted on the object is:

W = (15\,N)\cdot (100\,m)

W = 1500\,J

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object (a), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2} (2)

Where v_{o} is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

\frac{1}{2}\cdot a \cdot t^{2} =  \Delta x-v_{o}\cdot t

a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}

If we know that \Delta x = 100\,m, v_{o} = 0\,\frac{m}{s} and t = 10\,s, then the net acceleration experimented by the object is:

a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}

a = 2\,\frac{m}{s^{2}}

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

\Sigma F = F - f = m\cdot a (3)

Where:

F - External force exerted on the object, measured in newtons.

f - Kinetic friction force, measured in newtons.

If we know that F = 15\,N, m = 6\,kg and a = 2\,\frac{m}{s^{2}}, the kinetic friction force is:

f = F-m\cdot a

f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)

f = 3\,N

The work done by friction (W'), measured in joules, is:

W' = f\cdot \Delta x (4)

W' = (3\,N) \cdot (100\,m)

W' = 300\,J

And the net work experimented by the object is:

\Delta W = 1500\,J - 300\,J

\Delta W = 1200\,J

By the Work-Energy Theorem we understand that change in translational kinetic energy (\Delta K), measured in joules, is equal to the change in net work. That is:

\Delta K = \Delta W (5)

If we know that \Delta W = 1200\,J, then the change in translational kinetic energy is:

\Delta K = 1200\,J

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0
3 years ago
Light waves from the sun can be converted to electricity through __________.
ddd [48]

Answer:

light waves can be converted to electricity through <em>a solar cell</em>

Explanation:

3 0
3 years ago
A piece of copper wire with thin insulation, 200 m long and 1.00 mm in diameter, is wound onto a plastic tube to form a long sol
bearhunter [10]

Answer:

 N= 3

Explanation:

For this exercise we must use Faraday's law

          E = - dФ / dt

         Ф = B . A = B Acos θ

tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives

         E = - A cos θ (B - B₀) / t

The angle enters the magnetic field and the normal to the area is zero

         cos 0 = 1

         A = π r²

   

In the length of the wire there are N turns each with a length L₀ = 2π r

          L = N (2π r)

          r = L / 2π N

    we substitute

          A = L² / (4π N²)

The magnetic field produced by a solenoid is

           B = μ₀ N/L   I

for which

            B₀ = μ₀  N/L   I

           

The final field is zero, because the current is zero

            B = 0

We substitute

           E = - (L² / 4π N²)  (0 - μ₀ N/L I) / t

           E = μ₀ L I / (4π N t)

           N = μ₀ L I / (4π t E)

The electromotive force is E = 0.80 mV = 0.8 10⁻³ V

let's calculate

           N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]

           N  = 320 10⁻⁷ / 9.6 10⁻⁶

           N = 33.3 10⁻¹

          N= 3

           

7 0
2 years ago
You are driving your motorcycle in a circle of radius 75 m on wet pavement. what is the fastest you can go before you lose tract
stira [4]

On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s

Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.

Maximum velocity of a road with friction is given by the formula,

v = μRg

where, v is the maximum velocity

μ is the coefficient of static friction

R is the radius of the circle road

g is the acceleration due to gravity

Given,

μ = 0.20

R = 75m

g = 9.8m/s²

On substituting the given values in the above formula,

v = 0.20× 75 ×9.8

v = 147m/s

So, the Maximum velocity of the wet road is 147m/s.

Learn more about Velocity here, brainly.com/question/18084516

#SPJ4

3 0
1 year ago
A light ray travels in the +x direction and strikes a slanted surface with an angle of 62° between its normal and the ty axis. T
Elena-2011 [213]

Answer:

- 0.6

Explanation:

Given that angle between normal y axis is 62° so angle between  normal

and x axis will be 90- 62 = 28 °. Since incident ray is along x axis , 28 ° will be the angle between incident ray and normal ie it will be angle of incidence

Angle of incidence = 28 °

angle of reflection = 28°

Angle between incident ray and reflected ray = 28 + 28 = 56 °

Angle between x axis and reflected ray = 56 °

x component of reflected ray

= - cos 56 ( it will be towards - ve x axis. )

- 0.6

6 0
2 years ago
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