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2 years ago

What must you do to calculate a meaningful value of distance from the following equation?

Physics

2 answers:

ss7ja [257]2 years ago

5 0

Answer:

Convert all the times to either hours or seconds

Explanation:

One time unit is in hours, and the other is in seconds. In order to do the math, the units need to be the same. So you either need to convert hours to seconds, or seconds to hours.

devlian [24]2 years ago

4 0

Answer:

-Convert all the times to either hours or seconds.

Explanation:

Given,

$d= (0.90km) + (12.5km/h) \times (31.5s)$

While doing a mathematical calculation all the units should be same otherwise the result will not be correct. Here two units of time i.e. second and hour are being used. We will convert the second to hour by dividing it by 3600 as there are 3600 seconds in one hour. The distance will be,

$d= (0.90km) + (12.5km/h) \times (\frac{31.5}{3600}h)$

$d= 0.90 + 12.5 \times \frac{31.5}{3600}$

$d= 0.90 + \frac{393.75}{3600}$

$d= 0.90 + 0.109$

$d= 1.009 km$

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Question No 1 Find the voltage drop across 24 ohm resistor and current flowing through 22 ohm resistor in the given circuit as s

lord [1]

Answer:

8.25 V

Explanation:

We can ignore the 22Ω and 122Ω resistors at the bottom. Since there's a short across those bottom nodes, any current will go through the short, and none through those two resistors.

The 2Ω resistor and the 44Ω resistor are in parallel. The equivalent resistance is:

1 / (1 / (2Ω) + 1 / (44Ω)) = 1.913Ω

This resistance is in series with the 12Ω resistor. The equivalent resistance is:

1.913Ω + 12Ω = 13.913Ω

This resistance is in parallel with the 24Ω resistor. The equivalent resistance is:

1 / (1 / (13.913Ω) + 1 / (24Ω)) = 8.807Ω

Finally, this resistance is in series with the 4Ω resistor. The equivalent resistance of the circuit is:

8.807Ω + 4Ω = 12.807Ω

The current through the battery is:

12 V / 12.807Ω = 0.937 A

The voltage drop across the 4Ω resistor is:

(0.937 A) (4Ω) = 3.75 V

So the voltage between the bottom nodes and the top nodes is:

12 V − 3.75 V = 8.25 V

4 0

3 years ago

Myoncedyret

konstantin123 [22]

Answer:

i need help too

Explanation:

8 0

3 years ago

A man can jump 1.5 m on earth. calculate the approximate

Olegator [25]

Answer:

18 m

Explanation:

G = Gravitational constant

m = Mass of planet = $\rho V$

$\rho$ = Density of planet

V = Volume of planet assuming it is a sphere = $\dfrac{4}{3}\pi r^3$

r = Radius of planet

Acceleration due to gravity on a planet is given by

$g=\dfrac{Gm}{r^2}\\\Rightarrow g=\dfrac{G\rho V}{r^2}\\\Rightarrow g=\dfrac{G\rho \dfrac{4}{3}\pi r^3}{r^2}\\\Rightarrow g=\dfrac{4G\rho\pi r}{3}$

So,

$g\propto \rho r$

Density of other planet = $\rho_p=\dfrac{1}{4}\rho_e$

Radius of other planet = $r_p=\dfrac{1}{3}r_e$

$\dfrac{g_e}{g_p}=\dfrac{\rho_e r_e}{\rho_p r_p}\\\Rightarrow \dfrac{g_e}{g_p}=\dfrac{\rho_e r_e}{\dfrac{1}{4}\rho_e\times \dfrac{1}{3}r_e}\\\Rightarrow \dfrac{g_e}{g_p}=12\\\Rightarrow g_p=\dfrac{g_e}{12}\\\Rightarrow g_p=\dfrac{9.8}{12}$

Since the person is jumping up the acceleration due to gravity will be negative.

From kinematic equations we have

$v^2-u^2=2g_es\\\Rightarrow u^2=v^2-2g_es\\\Rightarrow u^2=0-2\times -9.8\times 1.5\\\Rightarrow u^2=2\times 9.8\times 1.5$

On the other planet

$v^2-u^2=2g_ps\\\Rightarrow s=\dfrac{v^2-u^2}{2g_p}\\\Rightarrow s=\dfrac{0-(2\times 9.8\times 1.5)}{2\times -\dfrac{9.8}{12}}\\\Rightarrow s=18\ \text{m}$