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3 years ago

List and describe the forces that act on a block as it moves down the ramp at constant speed.

Physics

1 answer:

Vitek1552 [10]3 years ago

7 0

gravity, inertia, friction, kinetic energy?

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A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s

Paraphin [41]

Answer:

$\omega = 0.016\,\frac{rad}{s}$

Explanation:

The rotation rate of the man is:

$\omega = \frac{v}{R}$

$\omega = \frac{0.80\,\frac{m}{s} }{5\,m}$

$\omega = 0.16\,\frac{rad}{s}$

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

$(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega$

The final angular speed is:

$\omega = 0.016\,\frac{rad}{s}$

3 0

3 years ago

What is the frequency of light for which the wavelength is 7.1 × 102 nm?

Brilliant_brown [7]

Answer:

Frequency, $f=4.22\times 10^{14}\ Hz$

Explanation:

Given that,

Wavelength of the light,

$\lambda=7.1\times 10^2\ nm=7.1\times 10^2\times 10^{-9}\ m\\\\\lambda=7.1\times 10^{-7}\ m$

We need to find the frequency of light. We know that light is an electromagnetic wave. It moves with the speed of light. So,

$c=f\lambda$

f is the frequency of light

$f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8\ m/s}{7.1\times 10^{-7}\ m}\\\\f=4.22\times 10^{14}\ Hz$

So, the frequency of light is $4.22\times 10^{14}\ Hz$ . Hence, this is the required solution.

5 0

4 years ago

The second Law of Thermodynamics states that: A. spontaneous processes are characterized by the overall conversion of order to d

Georgia [21]

Answer:

Spontaneous processes are characterized by the overall conversion of order to disorder.

Explanation:

The second law of thermodynamics states that: A spontaneous process occurs only if there is an increase in entropy of a system and its surroundings.

Entropy, S, is a measure of the randomness or disorder of a system. It is measured in J/Kmol.

The change in entropy, ∆S = ∆H/T

Where ∆H = change in enthalpy, T = Temperature in Kelvin.

For,

I. An endothermic reaction, ∆S = positive (that is, ∆S is greater than zero), there is an increase in entropy, therefore, the reaction is spontaneous.

II. An exothermic reaction, ∆S = negative (that is, ∆S is less than zero) there is a decrease in entropy, so, the reaction is non-spontaneous.

III. A system at equilibrium, ∆S = 0.

Then,

The standard change in entropy of a reaction, ∆So reaction , is the difference in the standard entropies between products and reactants:

∆So reaction = n ∆Soproducts - m ∆Soreactants

Where, = sigma = sum of,

∆ = delta = change in,

n and m = stoichiometric coefficients of the products and reactants respectively.

Furthermore, the entropy of the system and surroundings is referred to as the entropy of the universe.

∆Suniverse = ∆Ssurroundings + ∆Ssystem.

Processes leading to an increase in entropy include melting, heating, vaporization, dissolving.

8 0

3 years ago

Under which of the following circumstances do neap tides occur?

Yakvenalex [24]

I think <span>its C
hope it helps

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4 0

4 years ago

Read 2 more answers

In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact

OLga [1]

Answer:

v=3.73m/s, $a=3.35m/s^{2}$

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

$K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}$

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

$K_{A}+U_{Ag}=U_{Cs}+W_{Cf}$

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}$

so we can now solve this for k, which is the constant of the spring.

$k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}$

and now we can substitute the respective data:

$k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}$

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

$K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}$

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}$

and we can solve for the Velocity in B, so we get:

$V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}$

we can now input all the data directly, so we get:

$V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}$

which yields:

$V_{B}=3.73m/s$

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

$\sum F=ma$