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3 years ago

Una mochila tiene una masa de 5.9 Kg y se mueve con una aceleración de: 0.53 m/ seg2. ¿ Cuál es la fuerza que debe aplicar ?.

Physics

1 answer:

zhannawk [14.2K]3 years ago

8 0

Answer:

3,13 N

Explanation:

Del enunciado de la segunda ley del movimiento de Newton;

F = ma

Dónde;

F = Fuerza aplicada

m = masa del cuerpo

a = aceleración del cuerpo

Sustitución de valores;

F = 5,9 kg * 0,53 m / seg2 = 3,13 N

Por tanto, la fuerza aplicada es 3,13 N

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Assuming the acceleration due to gravity on the moon is exactly one-sixth of the acceleration due to gravity on Earth, what is t

uysha [10]

I have assumed a weight of 120 N on Earth.

Answer:

The object weighs 20 N on the moon

Explanation:

Weight

The weight of an object depends on the mass m of the object and the acceleration of gravity g of the place they are in.

The formula to calculate the weight is:

W = m.g

If g_e is the acceleration of gravity on Earth, and g_m is the acceleration of gravity on the moon, we know:

$g_m=1/6 g_e$

Dividing by ge:

$g_m/g_e = 1/6$

An object of weight We=120 N on planet Earth has a mass of:

$m = 120 / g_e$

Multiplying by gm:

$m.g_m=120 g_m/g_e$

Substituting the ratio of accelerations of gravity:

$m.g_m=120 * 1/6$

Since m.gm is the weight on the Moon Wm:

$W_m=20~N$

The object weighs 20 N on the moon

5 0

3 years ago

Is the following statement true or false? Nuclear fission occurs naturally in stars .

Mashcka [7]

True it can happen naturally trough radiation and decay of elements not only in earth but other planets too.

4 0

4 years ago

A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,

gavmur [86]

Answer:

a. ρ $_\beta=1.996J/m^3$

b. $U_E=9.445x10^{-15} J/m^3$

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

$i=14A$ , $d=2.5mm$ , $R=3.3$ Ω, $l=1 km$ , $E_o=8.85x10^{-12}F/m$ , $u_o=4*x10^{-7}H/m$

ρ $_\beta=\frac{\beta^2}{2*u_o}$

ρ $_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2$

ρ $_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}$

ρ $_\beta=1.996J/m^3$

b. The electric field can be find using the equation:

$U_E=\frac{1}{2}*E_o*E^2$

$E=(\frac{i*R}{l})^2$

$U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2$

$U_E=9.445x10^{-15} J/m^3$

4 0

4 years ago

In an experiment, 108 J of work was done on a closed system. During this phase of the experiment, 79 J of heat energy was added

amid [387]

Answer:

187 J

Explanation:

First Law of Thermodynamics :

ΔQ = ΔW + ΔU

ΔQ : Heat. If it added to system then positive and if it is rejected by system then negative.

ΔW : Work. If it done by the system then positive and if it is done on system then negative.

ΔU : Internal Energy. If it positive then temperature of system increased and if it is negative then temperature of system decreased.

ΔQ = 79 J

ΔW = - 108 J

ΔU = ?

substituting the value in the equation:

79 = -108 + ΔU

∴ ΔU = 187 J

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vlabodo [156]

Answer:

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3 years ago