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katovenus [111]

3 years ago

15

A rock dropped into a pond produces a wave that takes 11.3 s to reach the opposite shore, 26.5 m away. the distance between cons

ecutive crests of the wave is 3 m. what is the frequency of the wave? answer in units of hz.

1 answer:

Kay [80]3 years ago

6 0

The wave takes 11.3 s to cover a distance of 26.5 m, so its speed is:
$v= \frac{S}{t}= \frac{26.5 m}{11.3 s}=2.35 m/s$

The distance between two consecutive crests is 3 m, and this corresponds to the wavelength of the wave. To find its frequency, we can use the relationship between the speed v, the wavelength $\lambda$ and the frequency f:
$f= \frac{v}{\lambda}= \frac{2.35 m/s}{3 m}=0.78 Hz$

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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

A 0.155 kg arrow is shot upward

solniwko [45]

Answer:

2.43J

Explanation:

Given parameters:

Mass of the arrow = 0.155kg

Velocity = 31.4m /s

Unknown:

Kinetic energy when it leaves the bow = ?

Solution:

The kinetic energy of a body is the energy in motion of the body;

it can be derived using the expression below:

K.E = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

Solve for K.E;

K.E = $\frac{1}{2}$ x 0.155 x 31.4 = 2.43J

3 0

2 years ago

Read 2 more answers

One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor

Bess [88]

Answer:

a) The final velocity is 20 m/s when the large-mass object is the one moving initially.

b) The final velocity is 9.0 m/s when the small-mass object is the one moving initially.

Explanation:

The momentum of the system is calculated as the sum of the momenta of each object. Each momentum is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

Then, the momentum of the system is the following:

m1 · v1 + m2 · v2 = (m1 + m2) · v

Where:

m1 = mass of the bigger object.

v1 = velocity of the bigger object.

m2 = mass of the smaller object.

v2 = velocity of the smaller object.

v = final velocity of the two objects after the collision.

Solving the equation for the final velocity:

(m1 · v1 + m2 · v2)/ (m1 + m2) = v

a) Let´s calculate the final velocity when the bigger object is moving:

(7.1 kg · 29 m/s + 3.2 kg · 0)/(7.1 kg + 3.2 kg) = v

<u>v = 20 m/s</u>

b) When the smaller object is moving:

(7.1 kg · 0 m/s + 3.2 kg · 29 m/s) / (7.1 kg + 3.2 kg) = v

<u>v = 9.0 m/s</u>

7 0

3 years ago

Which one do I press guys?

taurus [48]

Answer:

The Nucleus...

Explanation:

E

8 0

3 years ago

Explain the difference between what Isaac Newton and Louis de Broglie would have to say about the momentum of a particle that is

pishuonlain [190]

Answer:

Explained

Explanation:

Newton would resort to the classical mechanics and say that the momentum of the particle that is moving with a constant velocity will be given by: momentum = mass x velocity

this approach will highlight the particle nature and will not be relativistic.

De-Broglie will say that the momentum of the particle is related to its associated matter wave and the relation between them is given by:

$p = \frac{h}{\lambda}$

where \lambda = wavelength of the matter wave associated to the particle, h = planck's constant

and $p = \gamma\times mv$

thus, this highlights the wave nature of the particle and is also relativistic.

6 0

3 years ago