Which points are most efficient for the utilization of resources on a production possibilities diagram?

Explain how scientists know that elephants and hyraxes are related. Be sure to include anatomical similarities as well as fossil

leonid [27]

<em>Hyraxes, elephants and dugongs are more closely related to one another than to any other living animal. Well, that's not exactly true. The closest surviving relative of the dugong is the manatee, a fresh-water, New-World version of itself. Apart from this, and the fact that there are two species of elephant and several of hyrax, it's absolutely true. Hyraxes, elephants and dugongs evolved from a single common ancestor.</em>

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8 0

2 years ago

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A varying force is given by F=Ae ^-kx, where x is the position;A and I are constants that have units of N and m^-1 , respectivel

Burka [1]

W = ∫ (x from 0.1 to +oo) F dx

= ∫ (x from 0.1 to +oo) A e^(-kx) dx

= A/k x [ - e^(-kx) ](between 0.1 and +oo)

= A/k x [ 0 + e^(-k * 0.1) ]

<span> = A/k x e^(-k/10) </span>

4 0

2 years ago

A proton and an electron are held in place on the x axis. The proton is at x = -d, while the electron is at x = +d. They are rel

Over [174]

The protons and electrons are held in place on the x axis.
The proton is at x = -d and the electron is at x = +d. They are released at the same time and the only force that affects movement is the electrostatic force that is applied on both subatomic particles. According to Newton's third law, the force Fpe exerted on protons by the electron is opposite in magnitude and direction to the force Fep exerted on the electron by the proton. That is, Fpe = - Fep. According to Newton's second law, this equation can be written as
Mp * ap = -Me * ae
where Mp and Me are the masses, and ap and ae are the accelerations of the proton and the electron, respectively. Since the mass of the electron is much smaller than the mass of the proton, in order for the equation above to hold, the acceleration of the electron at that moment must be considerably larger than the acceleration of the proton at that moment. Since electrons have much greater acceleration than protons, they achieve a faster rate than protons and therefore first reach the origin.

6 0

2 years ago

A spring (k = 802 N/m) is hanging from the ceiling of an elevator, and a 5.0-kg object is attached to the lower end. By how much

love history [14]

Answer:

0.00256 m

Explanation:

For this case, we should use the Hook Law for a spring:

F= -Kx , the negative sign indicates that the force that tries to “restore” the original status of the spring, and is against the force that causes the spring´s displacement

This is, the force F required to stretch an elastic object (for example, a metal spring), is directly proportional the extension “x” of the spring

“x” can be the extension or compression of the spring, and “K” is a constant (in units of N/m)

We also know that, according to Newton´s Law:

F=m*a

m= mass

a= acceleration

Then:

m*a= -Kx

Finally, the spring has an original length of L₀, so:

- If the spring is compressed, the final strength will be: L = L₀ – x

- If the spring is extended, the final strength will be: L = L₀ + x

According to the statement:

K = 802 N/m

a = 0.41 m/s2

m = 5 Kg

Then:

x =- (m*a)/K = (5x0.41)/802 = 0.00256 m (we do not consider the negative sign, due to above explanation: it only indicates that the restoration force is always against the force imposed on the spring )

So, if the spring has an original length of L₀, when the elevator is accelerating upwards, the spring will stretch from L₀ to (L₀ – 0.00256) m

6 0

2 years ago

A giant wall clock with diameter d rests vertically on the floor. The minute hand sticks out from the face of the clock, and its

Katyanochek1 [597]

Answer:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

Explanation:

We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.

The x-component of the length of the minute hand is:

$d_{x}(t)=dcos(\theta (t))$ (1)

d is the length of the minute hand (d=D/2)
D is the diameter of the clock
t is the time (min)

Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:

$\theta (t)=\omega *t$ (2)

Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:

$\omega=\frac{\Delta \theta}{\Delta t}=\frac{\theta_{f}-\theta_{i}}{t_{f}-t{i}}=\frac{\pi/2-0}{15-0}=\frac{\pi}{30}$

Now, let's put this value on (2)

$\theta (t)=\frac{\pi}{30}*t$

Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

I hope it helps you!

6 0

2 years ago