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svet-max [94.6K]
3 years ago
15

330 g of water at 55°c are poured into an 855 g aluminum container with an initial temperature of 10°

Physics
1 answer:
Olenka [21]3 years ago
3 0
The final temperature of the system is 32.5°
we know,  H = mcT 
where, H = Heat content of the body 
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry 

The net heat remains constant i.e. 
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x \frac{900}{4200} x (T - 10) 

⇒ 14,850 - 330T = 183.21T  - 1832 

⇒ - 513.21 T = - 16682

or T = 32.5°
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60 POINTS model how sound produced by a singer is recorded by a microphone and reproduced by a speaker (sound model) WITH RUBIC
Nikolay [14]

Answer:

Explanation: i don't know bro

6 0
3 years ago
A loop of area 0.250 m^2 is in a uniform 0.020 0-T magnetic field. If the flux through the loop is 3.83 × 10-3T· m2, what angle
dangina [55]

Answer:

40.0⁰

Explanation:

The formula for calculating the magnetic flux is expressed as:

\phi = BAcos\theta where:

\phi is the magnetic flux

B is the magnetic field

A is the cross sectional area

\theta is the angle that the normal to the plane of the loop make with the direction of the magnetic field.

Given

A = 0.250m²

B = 0.020T

\phi = 3.83 × 10⁻³T· m²

3.83 × 10⁻³ = 0.020*0.250cosθ

3.83 × 10⁻³ = 0.005cosθ

cosθ = 0.00383/0.005

cosθ = 0.766

θ = cos⁻¹0.766

θ = 40.0⁰

<em>Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰</em>

4 0
3 years ago
The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at
denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m  _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

8 0
4 years ago
How does an atom change if all of its electrons are removed?
Alex
<span>The atom becomes positively charged.

When you add electrons to a neutral atom, it is no longer a neutral atom, it has a negative change and is an anion. When you take away electrons from a neutral atom, it is no longer a neutral atom- it becomes a positive atom, and is a cation.</span>


6 0
3 years ago
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
4 years ago
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