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ziro4ka [17]
3 years ago
7

A kettle heats 1.75 kg of water. The specific latent heat of vaporisation of water is 3.34 x 106 J/kg. How much energy would be

needed to boil off the water? Write the answer out in full (i.e. not in standard form).
Physics
1 answer:
labwork [276]3 years ago
7 0

The amount of energy needed to boil off the water is 5,845,000 J

Explanation:

The amount of thermal energy needed to completely boil a certain amount of a liquid substance already at boiling temperature is given by

Q=m\lambda_v

where

m is the amount of the substance

\lambda_v is the specific latent heat of vaporization of the substance

In this problem, the substance is water, and its mass is

m = 1.75 kg

Its specific latent heat of vaporization is

\lambda_v = 3.34\cdot 10^6 J/kg

Assuming that the water is already at boiling temperature, therefore, the thermal energy needed to boil off the water is

Q=(1.75)(3.34\cdot 10^6)=5,845,000 J

Learn more about specific heat:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30 second time interval. We take th
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Let say for every 5 s of time interval the speed will remain constant

so it is given as

v(mi/h)   16    21    23    26    33    30     28

now we have to convert the speed into ft/s as it is given that 1 mi/h = 5280/3600 ft/s

so here we will have

v(ft/s)      23.5    30.8     33.73     38.13     48.4     44     41.1

now for each interval of 5 s we will have to find the distance cover for above interval of time

d = v \times t

d = (23.5 + 30.8 + 33.73 + 38.13 + 48.4 + 44 + 41.1) \times 5

d = 1298.1 ft

so here it will cover 1298.1 ft distance in 30 s interval of time

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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

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

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia
Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

When the tiger reaches the ground, y = 0:

0 = 6.00 m - 1/2 · 9.81 m/s² · t²

2 · (-6.00 m) / -9.81 m/s² = t²

t = 1.11 s

We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

x0 = 0

2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

v0 = 1.80 m/s

The initial speed of the tiger is 1.80 m/s

4 0
3 years ago
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