Please help me :( Thank you!

Jobisdone [24]

Trick question? In order to have kinetic energy, an object must be moving. Therefore, in this case, kinetic energy would be 0. If it were asking about potential energy, it would be a different story.

8 0

3 years ago

Read 2 more answers

A truck covered 2/7 of a journey at an average speed of 40

sertanlavr [38]

Answer:

The amount of time for the whole journey is 8 hours.

Explanation:

A truck covered 2/7 of a journey at an average speed of 40 mph. Representing 1 the total of the trip traveled, then the rest of the distance traveled is calculated as: $1-\frac{2}{7} =\frac{5}{7}$

So if the truck covered the remaining 200 miles at $\frac{2}{7}$ , this means that $\frac{5}{7}$ of the trip represents the 200 miles. So, to calculate the total distance traveled by the truck, you apply the following rule of three: if $\frac{5}{7}$ of the route represents 200 miles, the integer 1 (which represents the total of the route), how many miles are they?

$miles=\frac{1*200miles}{\frac{5}{7} } =\frac{7}{5} *200 miles$

miles= 280

So the total distance traveled is 280 miles. Since speed is the relationship between the space traveled by an object and the time used for it ( $speed=\frac{distance}{time}$ ), then if the average of the entire trip was 35 mph and the distance traveled 280 miles, the time is calculated as:

$time=\frac{distance}{speed}=\frac{280 miles}{35 mph}$

time= 8 h

 The amount of time for the whole journey is 8 hours.

6 0

3 years ago

Looking straight downward into a rain puddle whose surface is covered with a thin film of gasoline, you notice a swirling patter

Ivanshal [37]

Answer:

Explanation:

Point beneath you forms a beautiful iridescent green

refractive index of Gasoline $n=1.38$

Wavelength of Green light is $\lambda =540\ nm$

Here light first traverse from air(n=1) to gasoline , it reflects from front surface of gasoline(n=1.38) so it suffers a phase change. After this light reflect from rear surface of gasoline and there is a decrease in refractive index(n=1.38 to n=1.33), so there is no phase change occurs .

For constructive interference

$2t=(m+\frac{1}{2})\cdot \frac{\lambda }{n}$

here t= thickness of gasoline film

n=refractive index

for $m=0$

$t=\frac{\lambda }{4n}$

$t=\frac{540}{4\times 1.38}$

$t=97.82\approx 98\ nm$

4 0

3 years ago

A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other mov

Brums [2.3K]

Answer: 330.88 J

Explanation:

Given

Linear velocity of the ball, v = 17.1 m/s

Distance from the joint, d = 0.47 m

Moment of inertia, I = 0.5 kgm²

The rotational kinetic energy, KE(rot) of an object is given by

KE(rot) = 1/2Iw²

Also, the angular velocity is given

w = v/r

Firstly, we calculate the angular velocity. Since it's needed in calculating the Kinetic Energy

w = v/r

w = 17.1 / 0.47

w = 36.38 rad/s

Now, substituting the value of w, with the already given value of I in the equation, we have

KE(rot) = 1/2Iw²

KE(rot) = 1/2 * 0.5 * 36.38²

KE(rot) = 0.25 * 1323.5

KE(rot) = 330.88 J

6 0

4 years ago

(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i

murzikaleks [220]

Answer:

a) $t = -\frac{ln(2)}{k}$

b) See the proof below

$A(t) = A_o 2^{-\frac{t}{T}}$

c) $t = 3T \frac{ln(2)}{ln(2)}= 3T$

Explanation:

Part a

For this case we have the following differential equation:

$\frac{dA}{dt}= kA$

With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

And if we integrate both sides we got:

$ln |A|= kt + c_1$

Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

$\frac{1}{2} = e^{kt}$

Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

$ln(\frac{1}{8}) = -\frac{t}{T} ln(2)$

And if we solve for t we got:

$t = T \frac{ln(8)}{ln(2)}$

We can rewrite this expression like this:

$t = T \frac{ln(2^3)}{ln(2)}$

Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

8 0

3 years ago