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3 years ago

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A bumper has a non negligible mass. After a collision with an ideal spring, the spring is compressed to a distance of 90% of the

value of its distance when the bumper's mass was negligible. Give a reasonable explanation for this decrease.

1 answer:

Y_Kistochka [10]3 years ago

3 0

Deacceleration causing reverse of speed impact on the body of the frame causing reverse of shockwave.

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Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid ra

BabaBlast [244]

Explanation:

It is given that,

Length of the helicopter, l = 3.1 m

The helicopter rotates, the length of helicopter will become the radius of circular path, r = 3.1 m

Angular speed of the helicopter, $\omega=280\ rev/min=29.32\ rad/s$

(a) The centripetal acceleration in terms of angular velocity is given by :

$a_c=r\times \omega^2$

$a_c=3.1\times (29.32)^2$

$a_c=2664.95\ m/s^2$

(b) Let v is the linear speed of the tip. The relation between the linear and angular speed is given by :

$v=r\times \omega$

$v=3.1\times 29.32$

v = 90.89 m/s

$\dfrac{v_{tip}}{v_{sound}}=\dfrac{90.89}{340}=0.267$

Hence, this is the required solution.

4 0

3 years ago

A slingshot shoots a marble. It travels at an average speed of 15 m/s for 12 seconds before hitting it's target. How far does it

salantis [7]

Multiply m/s by 12 seconds to get meters by itself. 15(12)=180 m

6 0

3 years ago

The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.

<u>Solving for the amount remaining</u>

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

$y=14.2* (0.9172535661)^{(31.8)}$ $y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}$

$y=14.2* 0.0641450581$ $y=14.2*(0.5)^{3.962518068}$

$y=0.9108598257$ $y=14.2* 0.0641450581$

$y=0.9108598257$

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.

8 0

2 years ago

What is a consequence of the second law of thermodynamics?

marshall27 [118]

Answer:Increase in the entropy of the universe are the result of chemical reactions

Explanation:

The second law tells about the quality of energy such that any isolated system tends to become more disordered. Any natural Process occurring on its own is irreversible and entropy is increasing for that process.

The entropy of the Universe is increasing which is the result of chemical reactions. Any chemical reaction which increases the number of gas molecules will generate more entropy. Entropy in simple meaning is the order of randomness.

4 0

4 years ago

A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of

ANEK [815]

Answer:

The distance the piece travel in horizontally axis is

L=3.55m

Explanation:

$a=2 \frac{rev}{s^{2}} \\h=0.820m\\r = 0.125 m
\\d=150rev$

$d= 155 rev = 155(2\pi ) = 310\pi rad$

$a= 2.0 \frac{rev}{s^{2} } = 2.0(2\pi ) = 4.0\pi \frac{rev}{s^{2} }$

$d=d_{i}+vo*t+\frac{1}{2}*a*t^{2} \\ di=0\\vo=0\\d=\frac{1}{2}*a*t^{2}\\t=\sqrt{\frac{2*d}{a}}\\t=\sqrt{\frac{2*310 rad}{4\frac{rad}{s^{2}}}} \\t=12.449$

$w=a*t\\w=4\frac{rad}{s^{2}}*12.449s\\ w=49.79 \frac{rad}{s}$

Now the angular velocity is the blade speed so:

$V=w*r\\V=49.79 \frac{rad}{s}*0.175m\\V=8.7 \frac{m}{s}$

assuming no air friction effects affect blade piece:

time for blade piece to fall to floor

$t=\sqrt{\frac{2*h}{g}}\\t=\sqrt{\frac{2*0.820m}{9.8\frac{m}{s^{2} } }}\\t=0.409s$

Now is the same time the piece travel horizontally

$L=t*V\\L=0.409s*8.7\frac{m}{s}\\L=3.55m$

blade piece travels HORIZONTALLY = (24.5)(0.397) = 9.73 m ANS

6 0

4 years ago