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2 years ago

Plz help a smol bean out (no links btw)

Physics

1 answer:

Marina86 [1]2 years ago

8 0

Hey! If this is on big ideas get a app called “Slader” It has answers to all math problems just like up the book you use > save it > type in the section you are doing > and look for the problems you are doing.

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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.

3 0

3 years ago

Read 2 more answers

The element radon is at the opposite end of the range, with the lowest specific heat of all naturally occurring elements. Radon'

allochka39001 [22]

Explanation:

Q = mc∆T

= (0.34 kg)(94 J/kg-°C)(25°C)

= 799 J

5 0

3 years ago

The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th

Darya [45]

Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = $\frac{v^2}{r}$ [v = linear velocity, r = radius of circular path]

=> F = m $\frac{v^2}{r}$ ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m $\frac{v^2}{r}$ = q v B sin θ [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m $\frac{v^2}{r}$ = q v B sin 90°

=> m $\frac{v^2}{r}$ = q v B

Divide both side by v;

=> m $\frac{v}{r}$ = qB

Make v subject of the formula

v = $\frac{qBr}{m}$

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = $\frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}$

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = $\frac{1}{2}$ mv²

m = mass of proton

v = velocity of the proton as calculated above

K = $\frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )$

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

8 0

2 years ago

Calculate the force that the 4kg block exerts on the 10kg block

Kryger [21]

Acceleration=force/mass=28/(10+4)=2m/s^2

force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block

vf=vi+at
-10=20*28/14 * t
t=30/2=15sec

i hope this can help you.

8 0

2 years ago

X-Rays contradict to?<br> a)beta b) alpha.<br> C) gamma

Vesna [10]

<h3>X-Rays contradict to?</h3>

<h3>C. gamma </h3>

a type of penetrating electromagnetic radiation produced by the radioactive disintegration of atomic nuclei

3 0

2 years ago

Read 2 more answers