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2 years ago

8

In un lento processo di riscaldamento di 200 g di H2O da 60 gradi a 100 gradi evaporano 10 g di H2O. Assumendo per il calore lat

ente di vaporizzazione dell'H2O il valore costante di 540 cal/g, stabilire , con un calcolo di prima approssimazione, il calore totale che si deve fornire all'H2O per eseguire il riscaldamento
soluzione: Q=13 000 circa cal

1 answer:

Lyrx [107]2 years ago

7 0

https://www.khanacademy.org/science/physics/thermodynamics/specific-heat-and-heat-transfer/v/specific-heat-and-latent-leat-of-fusion-and-vaporization-2

https://www.khanacademy.org/science/physics/thermodynamics/specific-heat-and-heat-transfer/v/specific-heat-and-latent-leat-of-fusion-and-vaporization-2

This link might help

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Calculate the change in velocity of a 0.070kg tennis ball hit by Serena with a force of 140 N over 0.020 s

Brilliant_brown [7]

V=at and a=F/m

140/.070 = 2000m/s^2

2000*.020 = 40m/s

The ball’s velocity increased by 40m/s.

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2 years ago

LekaFEV [45]

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7 0

1 year ago

Larger animals use proportionately less energy than smaller animals; that is, it takes less energy per kg to power an elephant t

marin [14]

Answer:

Part a)

$Q = 952 cal/day$

Part b)

$P = 46 Watt$

Part c)

$\Delta P = 54 W$

Explanation:

As we know that 5000 kg African elephant requires 70,000 Cal for basic needs per day

so we will have

$m = 5000 kg$

$Q = 70,000 Cal$

so we have energy required per kg

$E = \frac{70,000}{5000}$

$E = 14 cal/kg$

Part a)

now we know that per kg the energy required will be same

so we have mass of the human is 68 kg

so energy required per day is given as

$Q = 68 \times 14$

$Q = 952 cal/day$

Part b)

Resting power is the rate of energy in Joule required per sec

so it is given as

$P = \frac{952 \times 4186}{24\times 3600}$

$P = 46 Watt$

Part c)

resting power given in the book is

$P' = 100 W$

so this is less than the power given

$\Delta P = 100 - 46$

$\Delta P = 54 W$

6 0

2 years ago

You are looking down on a N = 9 turn coil in a magnetic field B = 0.5 T which points directly down into the screen. If the diame

Novay_Z [31]

Answer:

The magnitude of the voltage is $2.27\times10^{-4}\ V$ and the direction of the current is clockwise.

Explanation:

Given that,

Number of turns = 9

Magnetic field = 0.5 T

Diameter = 3 cm

Time t = 0.14 s

We need to calculate the flux

Using formula of flux

$\phi=NAB$

Put the value into the formula

$\phi=9\times\pi\times(1.5\times10^{-2})^2\times0.5$

$\phi=0.003180$

We need to calculate the emf

Using formula of emf

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{0.003180}{0.14}$

$\epsilon =-0.000227\ V$

$\epsilon=-2.27\times10^{-4}\ V$

Negative sign shows the direction of current.

Hence, The magnitude of the voltage is $2.27\times10^{-4}\ V$ and the direction of the current is clockwise.

8 0

3 years ago

A 46.0-kg box is being pushed a distance of 8.80 m across the floor by a force P whose magnitude is 171 N. The force P is parall

dolphi86 [110]

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

$W_P=Px$

P: applied force = 171N

x: distance in which the for P is applied = 8.80m

you replace the values of P and x and obtain:

$W_P=(171N)(8.80m)=1504.8J$

(b) The work don by the friction force is:

$W_f=F_fx=\mu N x=\mu Mg x$

μ = coefficient of kinetic friction = 0.250

M: mass of the box = 46.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J$

(c) The Normal force is

$N=Mg=(46.0kg)(9,8m/s^2)=450.8N$

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

$W_N=0J$

(d) the same as before:

$W_g=0J$

8 0

2 years ago