True or False: Chemical energy stored in food cannot be transformed into mechanical energy

A truck is moving with a certain uniform velocity. It is accelerated uniformly by 0.75 m/s^2. After 20 seconds , the velocity be

scoundrel [369]

Answer:

Vi = 5 m/s

Explanation:

let (a) acceleration = 0.75 m/s²

(t) time = 20 seconds

Vf = final velocity = 72 km/hr (convert to m/s to units consistency = 20 m/s)

find Initial velocity (Vi)

Vf - Vi

a = -----------

t

Vi = Vf - (a * t) = 20 - (0.75 * 20)

Vi = 5 m/s

5 0

3 years ago

If a 4N weight is hung on a spring, and it extends by 0.2m, what is the spring constant (k)?

Naddika [18.5K]

Answer: 200 N/m

Explanation:

The Gravitational spring energy(Us) is equal to 1/2kx^2. So we have x as .2 m and Us as 4 N. So 4 N = 1/2 * k * .2^2. So now we solve for K and get 200 N/m.

4 0

3 years ago

Read 2 more answers

According to the "Law of Increasing Opportunity Costs," what would be the opportunity cost of a student who is staying up all ni

kodGreya [7K]

C. The opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

<h3>What is law of opportunity cost?</h3>

The law of increasing opportunity cost is an economic principle that describes how opportunity costs increase as resources are applied.

As the student gives up his sleep or night rest in the place of his exam preparation, we say that the opportunity cost is the sleep or rest.

Thus, the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

Learn more about opportunity cost here: brainly.com/question/8846809

#SPJ1

7 0

2 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

Three monkeys A, B, and C weighing 20, 26, and 25 lb, respectively, are climbing up and down the rope suspended from D. At the i

Marina86 [1]

Answer: $2235.2lb-ft/s^2$

Explanation:

Given

Mass of monkey A=20lb

Mass of monkey B=26lb

Mass of monkey C=25lb

acceleration of monkey A= $4.2ft/s^2$

acceleration of monkey B=0

acceleration of monkey C= $-5.4ft/s^2$

Force Due to monkey A $\left ( F_A\right )=20\times 4.2 =84lb-ft/s^2\left ( downwards\right )$

Force Due to monkey A $\left ( F_B\right )=26\times 0 =0lb-ft/s^2$

Force Due to monkey A $\left ( F_C\right )=25\times 5.4 =135 lb-ft/s^2\left ( upward\right )$

In addition to it Weights of monkeys will be acting downwards therefore net Downwards force is balanced by tension

T= $\left ( 20+26+25\right )32.2+84-135=2235.2 lb-ft/s^2$

5 0

3 years ago