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1 year ago

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En la etiqueta de un bote de fabada de 350 g, leemos que su aporte energético es de 1630 kj por cada 100 g de producto a) La can

tidad de fabada que tiene que comer un hombre si su necesidad energética diaria es 2500 kj para sobrevivir b) El aporte calórico del bote de fabada en calorías

1 answer:

fredd [130]1 year ago

6 0

Answer:

(a) 153.37 g

(b) 5705 kJ

Explanation:

(a) To find the amount of bean needed by a man you first calculate the equivalence in beans to 2500kJ

$2500kJ*\frac{100g}{1630kJ}=153.37\ g$

Thus, 153.37 g has the energy needed by a man that needs 200kJ per day.

(b) The amount of energy per pot of bean is given by:

$E=350g*\frac{1630kJ}{100g}\\\\E=5705\ kJ$

Thus, the energy is 5705kJ

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What charge accumulates on the plates of a 2.0-μF air-filled capacitor when it is charged until the potential difference across

enot [183]

Answer:

0.0002 C.

Explanation:

Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)

Mathematically, charge can be expressed as

Q = CV ................................. Equation 1

Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.

Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.

Substitute into equation 1

Q = 2×10⁻⁶× 100

Q = 2×10⁻⁴ C

Q = 0.0002 C.

The amount of charge accumulated = 0.0002 C

7 0

1 year ago

Read 2 more answers

The drawing shows three identical springs hanging from the ceiling. Nothing is attached to the first spring, whereas a 4.50-N bl

Pavlova-9 [17]

Answer:

a. 30 N / m

b. 9.0 N

Explanation:

Given that

Unstretched length of the spring, $L_o$ = 20.0cm = 0.2m

a) When the mass of 4.5N is hanging from the second spring, then extended length Is

$L_1$ = 35.0cm = 0.35m

So, the change in spring length when mass hangs is

$x = L_1 - L_o$

= (0.35 - 0.20) m

= 0.15m

As spring are identical

Let us assume that the spring constant be "k", so at equilibrium

Restoring Force on spring = Block weightage

kx = W = 4.50

$k= \frac{4.50}{x} = \frac{4.50}{0.15}$

= 30 N / m

b) Now for the third spring, stretched the length of spring is

$L_2$ = 50cm = 0.5m

So, the change in spring length is

$x'= L_2 - L_o$

= (0.5-0.20)m

= 0.30m

At equilibrium,

Restoring Force on spring = Block weightage

Now using all mentioned and computed values in above,

$W'= kx'$

= 30(0.3)

= 9.0 N

4 0

1 year ago

A transmission channel is made up of three sections. The first section introduces a loss of 16dB, the second an amplification (o

AlekseyPX

Answer:

$P_{out} = 0.100 W = 100 mW$

Explanation:

The attached image shows the system expressed in the question.

We can define an expression for the system.

The equivalent equation for the system would be

$G_{total} = G_{1} + G_{2} + G_{3}\\G_{total} = -16dB+20dB-10 dB = -6 dB$

so, the input signal could be expressed in dB terms

$P_{in} [dB] = 10 log_{10}(P_{in}) \\P_{in} [dB] = 10 log_{10}(0.4)\\P_{in} [dB] = -3.97 dB$ (1)

so the output signal could be expressed as.

$P_{out} = P_{in} + G_{1} + G_{2} + G_{3}\\P_{out} = -3.97 dB - 6dB = -9.97 dB$

The gain should be expressed in dB terms and power in dBm terms so

$P_{out} = -9.97 + 30 = 20.03 dBm$

using the (1) equation to find it in terms of Watts

$P_{out} = 0.100 W = 100 mW$

3 0

2 years ago

A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car p

DedPeter [7]

Answer:

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

Explanation:

<em><u>The knowable variables are </u></em>

$d_{t}=25m$

$t_{1}=1.3 s$

$t_{2}=3.9 s$

$t_{3}=5.5 s$

Since the three traffic signs are <u>equally spaced</u>, the <u>distance between each sign is $\frac{25}{3} m$ </u>

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

Since we know the velocity in two points and the time the car takes to pass the traffic signs

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

6 0

1 year ago

The length of a 60 W, 240 Ω light bulb filament is 60 cm Remembering that the current in the filament is proportional to the ele

faust18 [17]

Answer:

Finally current will be

i = 0.35 A

Explanation:

As we know that power of the bulb is given by the formula

$P = \frac{V^2}{R}$

now we have

$P = 60 W$

R = 240 ohm

so we have

$60 = \frac{V^2}{240}$

$V = 120 Volts$

now the current in the bulb is given as

$i = \frac{V}{R}$

$i = \frac{120}{240} = 0.5 A$

now when length of the filament is double

so the resistance of the wire also gets double

so we have

$P = \frac{V^2}{R}$

$60 = \frac{V^2}{480}$

$V = 169.7 volts$

now the current in the bulb is given as

$V = i R$

$169.7 = i(480)$

$i = 0.35 A$

8 0

1 year ago