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1 year ago

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A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a sma

ll pebble is caught in the tread of one of them. (A) What is the angular acceleration of the pebble during those two seconds

1 answer:

konstantin123 [22]1 year ago

8 0

Answer:

$a=2.5\ m/s^2$

Explanation:

Given that,

Initial speed, u = 5 m/s

Final speed, v = 10 m/s

Time, t = 2 s

The radius of the tire of the bike, r = 35 cm

We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.

$a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2$

So, the required angular acceleration of the pebble is equal to $2.5\ m/s^2$ .

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Where c is 3 x 10^8 ms^-1 and λ is the wave length

We also need the expression of connecting frequency to energy of photon

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Combining the two equations will give us:

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2 years ago

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The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviatio

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Answer:

2.25 %

Explanation:

65-95-99.7 is a rule to remember the precentages that lies around the mean.

at the range of mean ( $\mu$ ) plus or minus one standard deviation ( $\sigma$ ), $P([\mu-\sigma \leq X \leq \mu+\sigma])\approx 68.3%$

at the range of mean plus or minus two standard deviation, $P([\mu -2\sigma \leq X \leq \mu+2\sigma])\approx 95.5%$

at the range of mean plus or minus three standard deviation, $P([\mu - 3\sigma\leq X \leq \mu+3\sigma])\approx 99.7%$

So, note that they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) ( $P(X \leq \mu+2\sigma)$ )

So we know that the 95.5% is between $\mu - 2\sigma = 0.3 -2*0.1 = 0.28$ and $\mu + 2\sigma = 0.3 +2*0.1 = 0.32$ , hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than 0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.

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1 year ago

In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d

Alexxandr [17]

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

$x*g = x * \frac{dv}{dt} + v^2$

- Where, v = Speed of the raindrop

- Proposed solution to given ODE:

v = a*t

Where, a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

a = dv / dt ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

v = dx / dt , v = a*t ... proposed solution (Eq 2)

v .dt = dx = a*t . dt

- integrate both sides:

∫a*t . dt = ∫dt

x = 0.5*a*t^2 ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

= 1.5 a^2 t^2

a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

x (t) = 0.5*a*t^2

x (t = 3.0) = 0.5*9.81*3^2 / 3

x (3.0) = 14.7 m

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

m (t) = k*x (t)

m (3.0) = 2.00*x(3.0)

m (3.0) = 2.00*14.7

m (3.0) = 29.4 g

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2 years ago

What is the formula that can be used to find velocity if kinetic energy and mass are known?

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The formula for kinetic energy is $\frac{1}{2}m\Delta v^2$ . Thus, the equation for velocity is $v= \sqrt{ \frac{2TotalKineticEnergy}{m} }$ .

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1 year ago

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Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree tha

emmasim [6.3K]

Answer:

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Explanation:

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$\tan\theta=\dfrac{4}{15}$

$\tan\theta=0.266$

$\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}$

$\cos\theta=0.966$

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

$x=(u\cos\theta)t$

$t=\dfrac{x}{u\cos\theta}$

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

$y=u\sin\theta(t)-\dfrac{1}{2}gt^2$

$y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2$

$y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}$

Put the value into the formula

$y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})$

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$d=h-y$

$d=4-3.517$

$d=0.483\ m$

Hence, The distance between knothole and the paint ball is 0.483 m.

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2 years ago