If a lever has an input arm of 80 cm and an output arm of 20 cm, what is its ideal mechanical advantage?

A 200-N force acts on a 10-kg object. The acceleration of the object is

jeyben [28]

200N/10kg = 20 m/s^2

3 0

2 years ago

a ball is dropped from rest at a height of 89m above the ground. (a)what is it's speed just before it hits the ground? (b) how l

kykrilka [37]

Answer:

(a) 41.75m/s

(b) 4.26s

Explanation:

Let:

Distance, D = 89m

Gravity, $g$ = 9.8 m/ $s^{2}$

Initial Velocity, $u$ = 0m/s

Final Velocity, $v$ = ?

Time Taken, $t$ = ?

With the distance formula, which is

D = $ut$ + $\frac{1}{2} gt^2$

and by substituting what we already know, we have:

89 = $\frac{1}{2}$ ×9.8× $t^{2}$

With the equation above, we can solve for $t$ :

$t=\sqrt{\frac{89(2)}{9.8}} \\t=\sqrt{\frac{178}{9.8} } \\t=\sqrt{18.16} \\t=4.26 seconds$

Now that we have solved $t$ , we can use the following velocity formula to solve for $v$ :

$v = u + at$ , where $a$ is also equals to $g$ , so we have

$v = u + gt$

By substituting $u = 0$ , $g = 9.8$ , and $t = 4.26$ ,

We have:

$v = 0 + 9.8(4.26)\\v = 41.75m/s$

4 0

2 years ago

Plz, help me! I will give brainliest

Viktor [21]

It’s A liquid to a solid because if it’s a liquid they can move but not much and then it becomes a solid because they are packed closely together and are fixed in one position.

3 0

2 years ago

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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:

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

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

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:

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

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

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

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

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

2 years ago

A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of

Sholpan [36]

$F=ma$
where:
F - force
m - mass
a - acceleration

We transform this formula to get a:
$a= \frac{F}{m}$
$a=\frac{2}{10}\frac{N}{kg}=0.2\frac{m}{s^{2}}$

4 0

2 years ago

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