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aev [14]
3 years ago
12

the temperature of a body fell from 100°c to 50°c in 10 minutes. the surrounding temperature was 20°c. what is the temperature a

fter further 10 minutes​

Physics
1 answer:
Anna007 [38]3 years ago
4 0

Answer:

The temperature is 10 c

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Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2
Bess [88]

Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

      The mass of the second block is  M_2 = 5.45 \ kg

     The angle of inclination of the second block is  \theta _2 =  32.5^o

      The coefficient of kinetic friction of the second block is \mu _2 = 0.105

The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

substituting values a =  \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}

    => a =  0.7156 m/s^2

     

3 0
3 years ago
A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit
Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c =  3 * 10⁸ Hz

Wavelength of the incident wave in air:

\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3

\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

n_1 = \sqrt{\frac{4\pi * 10^{-7}  }{8.84 * 10^{-12}  } }

n_1 = 377 \Omega

The intrinsic impedance of media 2 is given as:

n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

ϵr = 9

n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }

n_2 = 125.68 \Omega

c) The reflection coefficient,r  and the transmission coefficient,t at the boundary.

Reflection coefficient, r = \frac{n - n_{0} }{n + n_{0} }

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

r = \frac{3 - n_{0} }{3 + n_{0} }

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is E_{0} = 10 V/m

Maximum amplitudes in the total field is given by:

E = tE_{0} and E = r E_{0}

E = 10r, E = 10t

3 0
3 years ago
Which of the below is an example of mimicry that enables prey species avoid predation?
Marta_Voda [28]

b..a harmless organism imitating the look of a harmful organism

Explanation:

A harmless organism imitating the look of a harmful organism is one example of mimicry that enables prey species avoid predation.

Prey are smaller and less harmful organisms often hunted by larger organisms usually carnivores.

Mimicry is a form of evolutionary adaptation process in which two organisms of the same specie or different species tends to look alike.

It is a subtle defense mechanism developed by organism over an extended period of time.

learn more:

Adaptation brainly.com/question/11105547

#learnwithBrainly

6 0
3 years ago
Hulk starts at -2 m, jumps to -15 m, and settles at 6 m. What is Hulk's displacement?
Lorico [155]

Answer:

Sorry don't know the answer

6 0
2 years ago
Which are uses of convex mirrors? Check all that apply.
Serga [27]

Answer:

Its B, C, E

Explanation:

Have a great day everyone:)

3 0
3 years ago
Read 2 more answers
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