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3 years ago

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Why is it that, in a spoon, you can see yourself upside down? Bamboozled me for years.

1 answer:

Delvig [45]3 years ago

6 0

Here is your answer.

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The formula is x = 1/2 at^2 and I have managed to fill in the variables as this. d = 1/2 9.81 m/s^2 1^2

Artyom0805 [142]

Right, as you mentioned in the comments, you find $d$ by plugging in the different values of $t$ .

For $t=1\,\mathrm s$ , we have

$d=\dfrac12\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(1\,\mathrm s)^2$

$d=\left(4.905\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(1\,\mathrm s^2\right)$

$d=4.905\,\mathrm m$

Similarly, for $t=2\,\mathrm s$ , you get

$d=\dfrac12\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(2\,\mathrm s\right)$

$d=\left(4.905\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(4\,\mathrm s^2\right)$

$d=19.62\,\mathrm m$

8 0

3 years ago

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential

castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge $q=+10.5\times10^{-6}\ C$

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot d$

$\Delta U=-q(E\cdot d)$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}$

$\Delta U=-3.32\times10^{-3}\ J$

The change in electric potential energy is $-3.32\times10^{-3}\ J$

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

$\Delta U=3.32\times10^{-3}\ J$

The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

3 0

3 years ago

To reduce inductive reactance, what devices are normally placed on transmission or distribution lines?

UkoKoshka [18]

Answer : Capacitors

Explanation : Capacitors are normally placed on transmission or distribution lines when to reduce inductive reactance.

This is because it enhances electromechanical and voltage stability , limit voltage dips at network nodes and reduces the power loss.

So, we can say that inductive reactance normally replace by the capacitors.

4 0

2 years ago

A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold

tino4ka555 [31]

$q_{c}$ = Heat released to cold reservoir

$q_{h}$ = Heat released to hot reservoir

$W_{max}$ = maximum amount of work

$t_{c}$ = temperature of cold reservoir

$t_{h}$ = temperature of hot reservoir

we know that

$\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}$

$q_{h} = (\frac{t_{h}}{t_{c}})q_{c}$ eq-1

maximum work is given as

$W_{max}$ = $q_{h}$ - $q_{c}$

using eq-1

$W_{max}$ = $(\frac{t_{h}}{t_{c}})q_{c}$ - $q_{c}$

6 0

3 years ago

A track star runs 100 meters in 10 seconds. What is the star's average speed?

kakasveta [241]

Answer:

10m/s

Explanation:

4 0

2 years ago