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Ivahew [28]
3 years ago
11

Absolute zero (K=0 or -273.15°C) is what temperature on the Farenheit scale? a) 459.4°F b) -301.43 °F c) 233.05°F d) -40.15°F e

) None of these is true
Physics
1 answer:
ANEK [815]3 years ago
3 0

Answer:

e) None of these is true

Explanation:

Given that

Temperature = 0 K

We know that relationship between kelvin and Farenheit scale

\dfrac{K-273}{100}=\dfrac{F-32}{180}

Now by putting the values

\dfrac{K-273}{100}=\dfrac{F-32}{180}

\dfrac{0-273}{100}=\dfrac{F-32}{180}

So F= - 459.67°F

So we can say that 0 K is equal to  - 459.67°F.

So the our option e is correct.

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A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm
I am Lyosha [343]

Answer:

Rate of change of magnetic field is 3.466\times 10^3T/sec        

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop r=\frac{0.13}{2}=0.065m

Length of the circular loop L=2\pi r=2\times 3.14\times 0.065=0.4082m

Wire is made up of diameter of 2.6 mm

So radius r=\frac{2.6}{2}=1.3mm=0.0013m

Cross sectional area of wire A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2

Resistivity of wire \rho =2.18\times 10^{-8}m

Resistance of wire R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm

Current is given i = 11 A

So emf  e=11\times 1.67\times 10^{-3}=0.0183volt

Emf induced in the coil is e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}

0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}

\frac{dB}{dt}=3.466\times 10^3=T/sec

8 0
3 years ago
After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. when the box reaches position
KiRa [710]

As we know by work energy theorem

total work done = change in kinetic energy

so here we can say that wok done on the box will be equal to the change in kinetic energy of the system

W_p = KE_f - KE_i

initial the box is at rest at position x = x1

so initial kinetic energy will be ZERO

at final position x = x2 final kinetic energy is given as

KE_f = \frac{1}{2}mv_1^2

now work done is given as

W_p = \frac{1}{2}mv_1^2 - 0

so we can say

W_p = \frac{1}{2}mv_1^2

so above is the work done on the box to slide it from x1 to x2

3 0
3 years ago
A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsul
jeka94

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

<h3>Given:</h3>

Mass of the astronaut, m_a = 126 kg

Speed he acquires, v_{a}  = 2.70 m/s

Mass of the space capsule, m_{c} = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

P_f = m_av_a + m_cv_c

P_I = 0

m_av_a + m_cv_c = 0

v_c =\frac{- m_a v_a}{m_c}}\\\\

   = \frac{126* 2.70}{1800}

   = - 0.189 m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2} × 126 × (2.70) ^2

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2}×1800×(0.189) ^2

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Learn more about kinetic energy here:

brainly.com/question/26520543

#SPJ1

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The word that identifies the color of an object seen under ordinary daylight is:
IRINA_888 [86]
The word that identifies the colour of an object seen under ordinary daylight is local colour.
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