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3 years ago

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If a Stone with an of original velocity of zero is falling from a ledge and takes eight seconds to hit the ground what is the fi

nal velocity of the stone

1 answer:

DaniilM [7]3 years ago

3 0

Let t=time to reach the ground=8 secs, g= acceleration of gravity. The speed v on reaching the ground is gt=8g=78.4 m/s where g=9.8 m/s/s approx.

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3) If an airplane is in flight cruising at constant velocity, what can be said about the net work

Crazy boy [7]

Explanation:

<h2>Yes!</h2>

<h3>In physics, constant velocity occurs when there is no net force acting on the object causing it to accelerate. In terms of airplane flight, the two main forces influencing its velocity forward are drag and thrust. At a constant altitude, when the force of thrust equals the opposing force of drag, then the airplane will experience uniform motion in one direction. This can be further explained by Newton’s First Law. </h3>

6 0

3 years ago

7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If th

Ainat [17]

Answer:

Explanation:

Given

mass of children $m_1=20\ kg$

$m_2=30\ kg$

distance between two children $L=3\ m$

suppose small child is at a distance of x m from pivot point

so torque of small child and heavier child must be equal

$20\times (x)=30\times (3-x)$

$2x=9-3x$

$5x=9$

$x=1.8\ m$

7 0

3 years ago

A stream moving with a speed of 7.1 m/s reaches a point where the cross-sectional area of the stream decreases to one half of th

lana66690 [7]

Answer:

14.2 m/s

Explanation:

Given data:

Speed of the stream, v₁ = 7.1 m/s

let the cross section area at initial point be A₁

now area at the second point, A₂ = (1/2)A₁ = 0.5A₁

now, from the continuity equation, we have

A₁v₁ = A₂v₂

where, v₂ is the velocity at the narrowed portion

thus, on substituting the values, we get

A₁ × 7.1 = 0.5A₁ × v₂

or

v₂ = 14.2 m/s

8 0

3 years ago

PLZ ANSWER ASAP IF YOU KNOW THIS...At the county fair, Carrie and Sam climbed up on the carousel horses. Around and around they

Maslowich

The answer is (a) because movement is acceleration

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2 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

<h3>a)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

<h3>b)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

7 0

3 years ago