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3 years ago

11

Al llegar a detenerse, un automóvil deja marcas de derrape de 92m de largo sobre una autopista. Si se supone una desaceleración

de 7 m/s, estime la rapidez del automóvil justo antes de frenar

1 answer:

True [87]3 years ago

4 0

Answer:

Explanation: no se español

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A patient is administered 20 mg of iodine-131. How much of this isotope will remain in the body after 40 days if the half-life f

igor_vitrenko [27]

The formula for half-life is:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}$

Where A is the amount of iodine-131 initially and after 40 days, t is time, h is half-life of the isotope. Let's plug in our values to the equation:

$A_{final}=20(\frac{1}{2})^{\frac{40}{8}=0.625g$

Therefore, the patient has 0.625 grams of iodine-131 after 40 days.

4 0

3 years ago

The average height of an apple tree is 4.00 meters. How long would it take an apple falling from that height to reach the ground

Elodia [21]

We know, s = ut + 1/2gt²
In free-fall, u = 0
so, t = √2s/g

Now, substitute the values,
t = √2*4 / 9.8 [ -ve sign shows opposite direction only ]
t = √8/ 9.8
t = 0.90 sec

In short, Your Answer would be 0.90 sec

Hope this helps!

8 0

4 years ago

Read 2 more answers

The man fire a 50-g arrow that moves at an unknown speed. It hits and embeds in a 350-g block that slides on an air track. At th

Natasha_Volkova [10]

Answer:

a) vAix = 80 m/s

b) The assumptions and implications were:

Assume that friction between the block and the surface it rests on does not change the momentum of the system during the collision.

Assume that friction is negligible throughout the process and the system’s internal energy does not change.

Assume all the system’s kinetic energy is converted into elastic potential energy at the end of the process.

If any of the assumptions are invalid then the arrow must have been travelling initially, vAix > 80 m/s

Explanation:

Arrow embeds into block

Take in the first instance the system to be the arrow + block (isolated). Establish reference coordinate system with the +x axis running horizontally in the direction of the arrow’s motion. The initial state (i) is the arrow travelling with velocity vAix and the final state (f) is the arrow embedded in the block. Now, apply the component form of the Generalized Impulse Momentum Equation to this system:

pAi + pBi + JonA + JonB = pAf + pBf

pAix + pBix + Jx = pAfx + pBfx

mA*vAix + mB*vBix + 0 = (mA + mB)*vfx

0.05*vAix + 0 = (0.05 + 0.35)*vfx

vAix = 8*vfx (1)

Arrow embeds into block

Now consider the next phase of motion and take as the system the arrow + block + spring. The initial state (i) is the arrow and block travelling with velocity equivalent to the final velocity from equation 1 (final state velocity in first phase becomes initial velocity in next phase); vix’ = vfx and the final state (f) is the arrow + block brought to rest and the spring compressed an amount, Δx = 0.1 m. Now, apply the Generalized Work Energy Principle to the system

Ei + W = Ef

Ki + Usi + W = Kf + Usf

0.5*(mA + mB)*vix’² = 0.5*k*Δx²

(0.05 + 0.35)*vfx² = 4000*(0.1)²

vfx = √(40/0.4) = 10 m/s

Substituting above back into equation 1:

vAix = 8* 10 m/s = 80 m/s

Arrow embeds into block

The assumptions and implications were:

Assume that friction between the block and the surface it rests on does not change the momentum of the system during the collision.

Assume that friction is negligible throughout the process and the system’s internal energy does not change.

Assume all the system’s kinetic energy is converted into elastic potential energy at the end of the process.

If any of the assumptions are invalid then the arrow must have been travelling initially, vAix > 80 m/s

7 0

3 years ago

A cylinder measuring wide and high is filled with gas. The piston is pushed down with a steady force measured to be . Calculate

jekas [21]

Explanation:

Let us assume that the given cylinder is 2.6 cm wide and its height is 3.1 cm. And, when piston is pushed down then the steady force is equal to 15 N.

Now, radius of the cylinder will be as follows.

r = $\frac{diameter}{2}$

= $\frac{2.6 cm}{2}$

= 1.3 cm

or, = 0.013 m (as 1 m = 100 cm)

As, area of cylinder = $\pi \times r^{2}$

= $3.414 \times (0.013 m)^{2}$

= $5.77 \times 10^{-4} m^{2}$

Relation between pressure and force is as follows.

Pressure = $\frac{Force}{Area}$

= $\frac{15 N}{5.77 \times 10^{-4} m^{2}}$

= 25996 $N/m^{2}$

Since, 1 $N/m^{2}$ = 1 Pa (as 1 kPa = 1000 Pa)

Therefore, P = 25996 $N/m^{2}$

= 25.99 kPa

= 26 kPa (approx)

Thus, we can conclude that pressure of the gas inside the cylinder is 26 kPa.

7 0

3 years ago

A 62.0-kg skier is moving at 6.30 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.90

Klio2033 [76]

Here are the missing questions:
(a) How fast is the skier moving when she gets to the bottom of the hill?
(b) How much internal energy was generated in crossing the rough patch?
Part A
The initial kinetic energy of the skier is:
$E_{k0}=m\frac{v_0^2}{2}$
Part of this energy is then used to do work against the force of friction. Force of friction on the horizontal surface can be calculated using following formula:
$F_f=mg\mu$
The work is simply the force times the length:
$W_f=F_f\cdot L=mg\mu L$
So when the skier passes over the rough patch its energy is:
$E=E_{k0}-W_f$
When the skier is going down the skill gravitational potential energy is transformed into the kinetic energy:
$E_p=E_{k1}\\ mgh=E_{k1}$
So the final energy of the skier is:
$E_f=E_{k0}-W_f+E_{k1}\\ E_f=m\frac{v_0^2}{2}-mg\mu L+mgh=1856.86$J$
This energy is the kinetic energy of the skier:
$E_f=m\frac{v_f^2}{2}\\ v_f=\sqrt{\frac{2E_f}{m}}=7.74\frac{m}{s}$
Part B
We know that skier lost some of its kinetic energy when crossing over the rough patch. This energy is equal to the work done by the skier against the force of friction.
$E_{int}=W_f\\
E_{int}=mg\mu L=894.1$J$

4 0

4 years ago