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Luda [366]
2 years ago
7

Calculate the calories lost when 95 g of water cools from 45 ∘C to 29 ∘C. Express your answer to two significant figures and inc

lude the appropriate units. Use abbreviations for the units.
Physics
1 answer:
lubasha [3.4K]2 years ago
8 0

Answer:

1,520.00 calories

Explanation:

Water molecules are linked by hydrogen bonds that require a lot of heat (energy) to break, which is released when the temperature drops. That energy is called specific heat or thermal capacity (ĉ) when it is enough to change the temperature of 1g of the substance (in this case water) by 1°C. Water ĉ equals 1 cal/(g.°C).

Given that ĉ = Q / (m.ΔT),

where Q= calories transferred between the system and its environment or another system (unity: calorie or cal) (what we are trying to find out),

m= mass of the substance (unity: grams or g), and

ΔT= difference of temperature (unity: Celsius degrees or °C); and

m= 95g and ΔT= 16°C:

Q= 1 cal/(g.°C).95g.16°C =<u> 1,520.00 cal </u>

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Someone plans to float a small, totally absorbing sphere 0.500 m above an isotropic point source of light,so that the upward rad
mote1985 [20]

Answer:

468449163762.0812 W

Explanation:

m = Mass = \rhoV

V = Volume =\dfrac{4}{3}\pi r^3

r = Distance of sphere from isotropic point source of light = 0.5 m

R = Radius of sphere = 2 mm

\rho = Density = 19 g/cm³

c = Speed of light = 3\times 10^8\ m/s

A = Area = \pi R^2

I = Intensity = \dfrac{P}{4\pi r^2}

g = Acceleration due to gravity = 9.81 m/s²

Force due to radiation is given by

F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}

According to the question

F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W

The power required of the light source is 468449163762.0812 W

4 0
2 years ago
Small frogs that are good jumpers are capable of remarkable accelerations. One species reaches a takeoff speed of3.7 m/s in60 ms
MAVERICK [17]

We know that acceleration is change in velocity by time taken for that change.

In this case velocity change is 3.7 m/s

Time taken for this change = 60 ms = 6 *10^{-3} seconds

So acceleration of frog  = \frac{3.7}{60*10^{-3}}

                                       = 61.66 m/s^2

So acceleration of frog is 61.66 m/s^2

o it is evident that frog is capable of remarkable accelerations.

8 0
2 years ago
Read 2 more answers
A block of size 20m x 10 mx 5 m exerts a force of 30N. Calculate the
Orlov [11]

Answer:

We know that force applied per unit area is called pressure.

Pressure = Force/ Area

When force is constant than pressure is inversely proportional to area.

1- Calculating the area of three face:

A1 = 20m x 10 m =200 Square meter

A2 = 10 mx 5 m = 50 Square meter

A3 = 20m x 5 m = 100 Square meter

Therefore A1 is maximum and A2 is minimum.

2- Calculate pressure:

P = F/ A1 = 30 / 200 = 0.15 Nm⁻²  ( minimum pressure)

P = F / A2 = 30 / 50 = 0.6 Nm⁻²   ( maximum pressure)

Hence greater the area less will be the pressure and vice versa.

3 0
2 years ago
What is the longest wavelength light capable of ionizing a hydrogen atom in the ground state?
Sindrei [870]

Answer:

9.12\cdot 10^{-8} m

Explanation:

The energy needed to ionize a hydrogen atom in the ground state is:

E=13.6 eV= 2.18\cdot 10^{-18}J

The energy of the photon is related to the wavelength by

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength

Solving the formula for the wavelength, we find

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{2.18\cdot 10^{-18}J}=9.12\cdot 10^{-8} m

7 0
2 years ago
Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50.
Nady [450]

Answer:

v = 38.73 m/s

Explanation:

Given

Extension of the bow, x = 50 cm = 0.5 m

Force of the arrow, F = 150 N

Mass of the arrow, m = 50 g = 0.05 kg

speed of arrow, v = ? m/s

We start by finding the spring constant

Remember, F = kx, so

k = F/x

k = 150 / 0.5

k = 300 N/m

the potential energy if the bow when pulled back is

E = 1/2kx²

E = 1/2 * 300 * 0.5²

E = 0.5 * 300 * 0.25

E = 37.5 J

The speed of the arrow will now be found by using the law of conservation of energy

1/2kx² = 1/2mv²

kx² = mv²

v² = kx²/m, on substituting, we have

v² = (300 * 0.5²) / 0.05

v² = 75 / 0.05

v² = 1500

v = √1500

v = 38.73 m/s

8 0
2 years ago
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