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3 years ago

Does anyone know how to work this out?

Physics

1 answer:

Vladimir [108]3 years ago

8 0

Answer:

$\frac{17}{14}*5$ = Approximately 6

Explanation:

What we need to know:

1. Given that this is a parallel circuit, each pathway will receive the full voltage from the source. Therefore, each pathway will receive 17V.

2. The current (I) stays the same for all loads in series. Therefore, I2 and I3 are equal since they are connected in series.

3. Ohm's law states that Voltage = Current × Resistance (V=IR)(I=V/R)(R=V/I)

1) Calculate the current for the path containing R2 and R3

$I_2_,_3=\frac{V_2_,_3}{R_2_,_3}\\I_2_,_3=\frac{17}{5+9}\\I_2_,_3=\frac{17}{14}$

Because the current running through this path is $\frac{17}{14}$ A, then I2 is $\frac{17}{14}$ A.

2) Use Ohm's Law by plugging in I2 and R2

$V_2=I_2*R_2\\V_2=\frac{17}{14} *5$

$V_2=6$ (approximate)

Therefore, V2 is approximately 6V.

I hope this helps!

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A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
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v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
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b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
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$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

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3 years ago

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rodikova [14]

Answer:

The gravitational field strength on the surface of Mars = 3.72 m/s²

Explanation:

Gravitational Field Strength: This can be defined as the force per unit mass which is exerted at that point. its direction is the force exerted on a mass in a gravitational field. The S.I unit of gravitational field strength is m/s²

Mathematically, Gravitational field is represented as,

g = F/m ..................... Equation 1.

m = F/g ..................... Equation 2.

Where g = gravitational Field Strength, F = force on the mass, m = mass of the body.

From the question,

Note: That The mass of the object is constant both on the surface of the earth and on the surface of Mars.

On the Surface of the earth,

Given: F = 785 N, g = 9.8 m/s²

Substituting this values into equation 2,

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m = 80.10 kg.

On the surface of Mars.

Given: m = 80.10 kg, F = 298.

Substituting into equation 2

g = 298/80.1

g = 3.72 m/s²

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