If the ratio between the frequency of a man's voice and that of a girl's voice is 3/4, what

A car traveling at 37m/s starts to decelerate steadily. It comes to a complete stop in 15 seconds. What is it’s acceleration

LuckyWell [14K]

Answer

The acceleration is

$a=-2.5ms^{-2}$ to the nearest tenth

Explanation

Since the car was travelling at $37ms^{-1}$ before it starts to decelerate, the initial velocity is

$u=37ms^{-1}$ .

The final velocity is $v=0ms^{-1}$ , because the car came to a stop.

The time taken is $t=15s$ .

Using the Newton's equation of linear motion,

$v=u +at$ , we find the acceleration by substituting the known values.

This implies that,

$0=37 +a(15)$

This gives us,

$0-37=15a$

$\Rightarrow -37=15a$

We divide both sides by 15 to get,

$a=-\frac{37}{15}ms^{-2}$

or

$a=-2.46667ms^{-2}$

6 0

3 years ago

Answer the three questions below that pertain to atomic orbitals. Part A: No two electrons can have the same set of quantum numb

zimovet [89]

Answer:

Pauli exclusion principle

Explanation:

No two fermions can have the same set of quantum numbers is called the Pauli exclusion principle. And a electron is a fermion.

There are four basic quantum numbers

Principle quantum number n

the angular momentum quantum number l

magnetic quantum number m_l

and electron spin quantum number m_s

8 0

3 years ago

What is a synonym & an antonym for property ? What is a synonym of physical change ?

tigry1 [53]

Answer:

A synonym for property is a quality or a characteristic. An antonym is non-physical. A synonym of physical change is a change in state or activity. Phase transition or Natural action.

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:

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