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ludmilkaskok [199]

2 years ago

10

Un hombre de pie puede ejercer la misma fuerza con sus piernas tanto en la tierra como en la luna. Sabemos que la masa del hombr

e es la misma en la Tierra y en la Luna. También sabemos que F=m x a (masa x aceleración, corresponde a la 2° Ley de Newton), es verdad tanto en la Tierra y la Luna. ¿Será el hombre capaz de saltar más alto en la Luna que en la Tierra? ¿Por qué o por qué no?

1 answer:

Paul [167]2 years ago

6 0

Answer:

The height jumped by the person on the moon is 6 times the height jumped by the person on earth.

Explanation:

As we know that the acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth.

So, it is false.

Let the mass of man is m and the gravity on moon is g' = g/6.

Let the height jumped on earth is h and the height jumped on moon is h'.

So,

m x g' x h' = m x g x h

g/6 x h' = g x h

h' = 6 h

So, the height jumped by the person is 6 times the height jumped by the person on earth.

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A carbon rod with a radius of 1.9 mm is used to make a resistor. What length of the carbon rod should be used to make a 3.7 Ω re

vladimir2022 [97]

Answer:

Length = 2.32 m

Explanation:

Let the length required be 'L'.

Given:

Resistance of the resistor (R) = 3.7 Ω

Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]

Resistivity of the material of rod (ρ) = $1.8\times 10^{-5}\ \Omega\cdot m$

First, let us find the area of the circular rod.

Area is given as:

$A=\pi r^2=3.14\times (0.0019)^2=1.13\times 10^{-5}\ m^2$

Now, the resistance of the material is given by the formula:

$R=\rho( \frac{L}{A})$

Express this in terms of 'L'. This gives,

$\rho\times L=R\times A\\\\L=\frac{R\times A}{\rho}$

Now, plug in the given values and solve for length 'L'. This gives,

$L=\frac{3.7\ \Omega\times 1.13\times 10^{-5}\ m^2}{1.8\times 10^{-5}\ \Omega\cdot m}\\\\L=\frac{4.181}{1.8}=2.32\ m$

Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.

5 0

3 years ago

To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.

ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current $i_1$ . In the case of resistance 2, it will cross a current $i_2$

Defined this we proceed to obtain our equations,

For $i_1$ ,

$\frac{di_1}{dt}+i_1R_1 = V$

$I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

For $i_2$ ,

$I_2R_2 =V$

$I_2 = \frac{V}{R_2}$

The current in the entire battery is equivalent to,

$i_t = I_1+I_2$

$i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

Our values are,

$V=1V$

$R_1 = 95\Omega$

$L= 1.5*10^{-2}H$

$R_2 =360\Omega$

Replacing in the current for t= 0.4m/s

$i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})$

$i= 0.0133A$

$i_1 = 0.01052A$

3 0

2 years ago

What is the momentum of a .005kg bumble bee that is traveling at a velocity of 3.0m/s?

stiks02 [169]

p=mv

p=0.005kg×3.0m/s

p= 0.015kgm/s

4 0

2 years ago

If the pressure in a gas is doubled while its volume is held constant, by what factor do vrms change

Nat2105 [25]

Answer is given below

Explanation:

given data

pressure = double

volume = constant

solution

As we know that an Average velocity and rms velocity is directly proportional to square root of PV ..................1

so if we take P is doubled while keeping V constant

than Velocity increases by a factor $\sqrt{2}$

so that Factor = 1.414 for both the cases

8 0

3 years ago

Light travels at the speed of approximately 3.0 × 108 meters per second. Find the time in minutes required for light to travel f

Solnce55 [7]

Answer:

t = 8.33 minutes

Explanation:

given,

Speed of Light, v = 3 x 10⁸ m/s

distance between sun to Earth = 1.5 x 10¹¹ m

time taken to reach earth = ?

we know,

Distance = speed x time

$t = \dfrac{d}{s}$

$t = \dfrac{1.5\times 10^{11}}{3\times 10^8}$

t = 500 s

1 min = 60 s

now, $t = \dfrac{500}{60}\ minutes$

t = 8.33 minutes

time taken by the light to reach earth is equal to 8.33 minutes.

3 0

3 years ago