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3 years ago

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A particle of mas 0.1 kg moves along the trajectory x(t)-3t t where x is in meters and t is in seconds. What is the kinetic ener

gy of the particle at time t? Select one: a. 0.3 (t 1)3 o b. 0.25 (t2 -1) c. 0.25 (t + 1)2 d. 0.45 (t2+1)2 o e. 3.0 (1-)

1 answer:

devlian [24]3 years ago

6 0

Answer:

Kinetic energy = 0.45 joules

Explanation:

We have velocity of the particle is

$v=\frac{dx(t)}{dt}\\\\v=\frac{d(3t)}{dt}\\v=3m/s$

Now kinetic energy(K.E) is given by

$K.E=\frac{1}{2}mv^{2}$

applying values we get

Kinetic energy = $\frac{1}{2}0.1\times 3^{2}\\\\\therefore K.E=0.45Joules$

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3) The work done is D. zero

4) The kinetic energy is B. 180 J

5) The potential energy is A. 120 J

6) The work done depends on B. position

7) The example of non-renewable energy is C. coal

8) The power expended is $3\cdot 10^4 W$

9) The efficiency is A. 100%

10) The velocity ratio is 5

Explanation:

3)

The work done by a force acting an object is given by:

$W=Fd cos \theta$

where :

F is the magnitude of the force

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When the force is applied perpendicular to the direction of motion,

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Therefore, the work done is:

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4)

The kinetic energy of a body is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the body

v is its speed

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v = 3 m/s

Therefore her kinetic energy is

$K=\frac{1}{2}(40)(3)^2=180 J$

5)

The potential energy of an object is given by

$PE=mgh$

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m is the mass

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h is the heigth of the object relative to the ground

For the ball in this problem,

m = 0.4 kg

h = 30 m

So, the potential energy is

$PE=(0.4)(10)(30)=120 J$

6)

A conservative field is a field for which the work done by the field on an object does not depend on the path taken, but only on the initial and final position of the object.

Gravitational and electric fields are examples of conservative fields. In fact:

When an object is pulled down by gravity (free fall), the work done by the gravitational field only depends on the change in height $\Delta h$ between the two points, not on the path taken during the fall
When an electric charge is pushed by the electric field, the work done by the field depends only on the initial and final position of the charge in the field

For any conservative field, it is possible to define a "potential" function, which represents the energy per unit mass/charge, and depends only on the position of the object.

7.

Non-renewable energy sources are sources of energy whose rate of consumption is faster than the rate at which they are re-created. Examples of non-renewable sources are coal, oil, natural gas. These energy sources are consumed at a fast rate, while they take million of years to regenerate, so at the current rate they will eventually run out.
Renewable energy sources are sources of energy that replenish at faster rate than the rate at which it is consumed. Examples of renewable sources are solar energy, wind, hydroelectric power.

Therefore, the example of non-renewable energy in this case is

C. Coal

8.

For an object pushed by a force F and moving at a constant velocity v, the power expended is given by

$P=Fv$

where F is the force and v is the velocity.

for the rocket in this problem, we have:

F = 10 N is the force propelling the rocket

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Substituting into the equation, we find the power expended:

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9.

The efficiency of a machine is given by

$\eta = \frac{W_{out}}{W_{in}}$

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$W_{in}$ is the energy in input to the machine

$W_{out}$ is the useful work in output from the machine

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which means 100%.

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The velocity ratio of a block and tackle system is the ratio between the distance moved by the effort and the distance moved by the load.

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In a block and tackle system, the velocity ratio is also equal to the number of pulleys in the system.

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jenyasd209 [6]

The coefficient of friction is missing and it has a value of μ = 0.4

Answer:

a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

Where;

N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

μ(N1 + N2) = (mp + mv) x a

Thus,

0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

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a = 4512.6/1150

a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

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