The half-life of iodine-131 is 8.04 days. It’s decay constant equals

A stone is thrown vertically upward with a speed of 18.0 . (a)How fast is it moving when it reaches a height of 11.0 ? (b)How lo

aliina [53]

For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative.
So Vf = square root [2(-9.8)(11.0) + (18.0)^2] 
Vf = 10.4 m/s your answer is correct.

For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, 11= 18t m/s-(1/2) 9.8t^2 then -11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s

5 0

3 years ago

A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans

alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

y = $(0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)$

Now, we will compare above equation with the standard form of transeverse wave equation,

y = $A sin(kx + \omega t)$

where, A is the amplitude = 0.09 m

k is the wave vector = $\frac{\pi}{11}$

$\omega$ is the angular frequency = $4\pi$

x is displacement = 1.40 m

t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave will be:

v(t) = $\frac{dy}{dt}$

v(t) = $A \omega cos(kx + \omega t)$

v(t) = $(0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)$

v(t) = -0.84 m/s

The acceleration of the particle in the location is

a(t) = $\frac{dv}{dt}$

a(t) = $-A \omega 2sin(kx + \omega t)$

a(t) = $-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)$

a(t) = -9.49 $m/s^{2}$

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 $m/s^{2}$ .

(b) Wavelength of the wave is given as follows.

$\lambda = \frac{2\pi}{k}$

$\lambda = (frac{2\pi}{\frac{\pi}{11})
$

$\lambda$ = 22 m

The period of the wave is

T = $\frac{2 \pi}{\omega}$

T = $\frac{2 \pi}{4 \pi}$

= 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

v = $\frac{\lambda}{T}$

= $\frac{22 m}{0.5 s}$

= 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

7 0

3 years ago

Car a runs a red light and broadsides car b, which is stopped and waiting to make a left turn. car a has a mass of 1,800kg. car

frez [133]

The law of conservation of momentum tells us that momentum is conserved, therefore total initial momentum should be equal to total final momentum. In this case, we can expressed this mathematically as:

mA vA + mB vB = m v

where, m is the mass in kg, v is the velocity in m/s

since m is the total mass, m = mA + mB, we can write the equation as:

mA vA + mB vB = (mA + mB) v

furthermore, car B was at a stop signal therefore vB = 0, hence

mA vA + 0 = (mA + mB) v

1800 (vA) = (1800 + 1500) (7.1 m/s)

vA = 13.02 m/s

7 0

3 years ago

4 This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive

zubka84 [21]

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

|

B1 at 20km/h

|

V

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

3 0

3 years ago

8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.

mart [117]

Answer:

The resultant tension of the two ropes is approximately 42.4 N

The length of the line representing the resultant tension is approximately 8.48 cm

Please find included with the answer the scale drawing created with Microsoft Word

Explanation:

The given parameters are;

The tension in rope P, $T_P$ = 30 N