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Gala2k [10]
3 years ago
15

If the work done to stretch an ideal spring by 4.0 cm is 6.0 J, what is the spring constant (force constant) of this spring? A)

300 N/m B) 3000 N/m C) 3500 N/m D) 7500 N/m E) 6000 N/m
Physics
1 answer:
Ostrovityanka [42]3 years ago
6 0

Answer:

D) 7500 N/m

Explanation:

Given :

Work done by the spring, W = 6 J

Stretch in the spring, x = 4 cm = 0.04 m

Consider k be the spring constant.

The relation between work done and stretch of the spring is:

W=\frac{1}{2}kx^{2}

Substitute the suitable values in the above equation.

6=\frac{1}{2}k(0.04)^{2}

k=\frac{12}{1.6\times10^{-3} }

k = 7500 N/m

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At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There
katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

\tau_{net} = I \alpha (1)

where

\tau_{net} is the net torque acting on the object in rotation

I is the moment of inertia of the object

\alpha is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

\alpha = \frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time interval

So we can rewrite eq.(1) as

\tau_{net}=I\frac{\Delta \omega}{\Delta t}

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

\tau_{net}=0

From the previous equation, this implies that

\Delta \omega =0

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

\tau_{net}=0

Therefore, this means that

\Delta \omega =0

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0
3 years ago
A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross
Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

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Our values are:

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C_d=0.45

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\rho=1.2kg/m^3

Replacing,

F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2

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We need calculate now the energy lost through a time T, then,

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But we know that d is equal to

d=vt

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v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

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W=87.47*10^6J (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

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Answer:

In conductive materials, the outer electrons in each atom can easily come or go, and are called free electrons. In insulating materials, the outer electrons are not so free to move. All metals are electrically conductive.

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D. none of them po

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