An object possesses both potential (PE) and kinetic (KE) energy. The object’s total mechanical energy is

Two ropes are attached to a 35 kg object. The first rope applies a force of 20 N and the second applies a force of 55 N. If the

Goshia [24]

Answer:

$a=1.672\ m.s^{-2}$

Explanation:

Given:

mass of the object, $m=35\ kg$

forces by two mutually perpendicular ropes of the attached to the object:

$F_x=20\ N$

$F_y=55\ N$

<u>Now we find the resultant force effect due to the two given forces:</u>

$F=\sqrt{F_x^2+F_y^2}$

$F=\sqrt{(20)^2+(55)^2}$

$F=58.52\ N$

<u>Now the acceleration will be due to this resultant force:</u>

$a=\frac{F}{m}$

$a=\frac{58.52}{35}$

$a=1.672\ m.s^{-2}$

3 0

3 years ago

Read 2 more answers

Identify the physical mechanism that causes turbulent thermal conductivity.

myrzilka [38]

Answer:

Option A is correct.

Eddies due to enhanced mixing of ﬂuid

Explanation:

Turbulent thermal conductivity is thermal conductivity that arises from the turbulent flow of fluids. It comes into play when a particukar fluid moves into turbulent regiom of flow where flow is no longer orderly and streamlines aren't discernable with the fluid layers all warping into one another forming vortices.

It is represented as K and is shown mathematically through the heat flux at turbulent flow

q = vCρT' = - K (∂T/∂y)

where

K = turbulent thermal conductivity

T' = the eddy temperature relative to the mean value,

C = Heat capacity the fluid

q = the rate of thermal energy transport by turbulent eddies.

The physical mechanism that cause turbulent thermal conductivity are similar to the causes of turbulent flow of fluids.

This includes sharp changes in fluid pressure and velocity of flow which is evident in eddies that come about in the enhanced mixing of fluids.

Hope this Helps!!!

5 0

3 years ago

When is thermal equilibrium achieved between two objects

melisa1 [442]

Answer:

when the temperatures of the two objects are equal

Explanation:

Thermal equilibrium is achieved between two objects when the temperatures of the two objects are equal. Heat flows from hot to cold objects. When the two objects attain equal temperatures, we say thermal equilibrium has been achieved

8 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$