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3 years ago

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Electrical Safety: What should you do to a hot plate before turning it on? More than one answer may be correct. Check that the l

ead does not trail in liquid. Check the cable for exposed wires. Position in a stable place away from places where it could be knocked over. Wipe any oil residues off the hot plate.

1 answer:

solniwko [45]3 years ago

8 0

Answer and Explanation:

Before turning a hot plate on, one should consider these points:

Check for the leads and ensure they does not trail in liquid.
The cable should be checked for any exposed wire before turning on in order to prevent any shock.
Any oil residue should be wiped off of the hot plate to ensure its longevity.
It should be placed in a stable place in order to ensure safety against any hazard.
Thus all the above mentioned points are correct to ensure safety.

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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:

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

and

F cos θ = f

when substituting one into the other we get:

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

which can be solved for the static friction coefficient so we get:

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

which is the answer to part b.

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3 years ago

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Which of the following never cause a change in the motion of an object? A. Net forces B. Unbalanced forces O C. Balanced forces

Karolina [17]

Answer:

balanced force never changes its motion

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2 years ago

One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force f is applied to the other end of th

Vlada [557]

For rotational equilibrium of the door we can say that torque due to weight of the door must be counter balanced by the torque of external force

$F\times L = mg \times \frac{L}{2}$

here weight will act at mid point of door so its distance is half of the total distance where force is applied

here we know that

$mg = 145 N$

now we will have

$F = \frac{mg}{2}$

$F = \frac{145}{2} = 72.5 N$

so our applied force is 72.5 N

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3 years ago

What describes how fast an object is moving and in what direction

Mashcka [7]

Speed is a description of how fast an object moves; velocity is how fast and in what direction it moves. In physics, velocity is speed in a given direction. When we say a car travels at 60 km/h, we are specifying its speed.

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Earth exerts a 100 N gravitational force on a metal box. What is the magnitude of the gravitational force the metal box exerts o

jeka94

We have that the magnitude of the gravitational force is mathematically given as

f=6.377N

<h3>Force</h3>

Question Parameters:

Earth exerts a 100 N gravitational force on a metal box.

(Mass of the earth is 6e24 kg and radius of the earth is 6.4e6m.)

Generally the equation for the Gravitational mForce is mathematically given as

$F=\frac{GMm}{r^2}\\\\Therefore\\\\F=\frac{100/9.8* 6e116e24}{6.4e6^2}$

f=6.377N

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2 years ago