What nuclear weapon is the biggest and who currently owns this bomb

On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4

anygoal [31]

Answer:

a) $\Delta s=443037.9747\ J.K^{-1}$

b) $\Delta s=31868131.8681\ J.K^{-1}$

Explanation:

a)

Given:

amount of heat transfer occurred, $dQ=3.5\times 10^6\ J$

initial temperature of car, $T_i=36.5+273=309.5\ K$

final temperature of car, $T_f=44.4+273=317.4\ K$

We know that the change in entropy is given by:

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)}$ (heat is transferred into the system of car)

$\Delta s=443037.9747\ J.K^{-1}$

b)

amount of heat transfer form the system of house, $dQ=5.8\times 10^8\ J$

initial temperature of house, $T_i=23.5+273=296.5\ K$

final temperature of house, $T_f=5.3+273=278.3\ K$

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{5.8\times 10^8}{278.3-296.5}$

$\Delta s=31868131.8681\ J.K^{-1}$

6 0

3 years ago

A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5\times 10^{-

joja [24]

Answer:

$5.25\cdot 10^{40} kg m^2/s$

Explanation:

The angular momentum of the pulsar is given by:

$L=m\omega r^2$

where

$m=2.8\cdot 10^{30} kg$ is the mass of the pulsar

$r = 10.0 km = 1\cdot 10^4 m$ is the radius

$\omega$ is the angular speed

Given the period of the pulsar, $T=33.5\cdot 10^{-3} s$ , the angular speed is given by

$\omega=\frac{2\pi}{T}=\frac{2 \pi}{33.5\cdot 10^{-3}s}=187.5 rad/s$

And so, the angular momentum is

$L=m\omega r^2=(2.8\cdot 10^{30}kg)(187.5 rad/s)(1\cdot 10^4 m)^2=5.25\cdot 10^{40} kg m^2/s$

8 0

3 years ago

How many signifficant figures do the following number have 258

Romashka [77]

Answer:

Facts about 258

Sig Figs

3

258

Decimals

0

Scientific Notation

2.58 × 102

E-Notation

2.58e+2

Words

two hundred fifty-eight

Explanation:

258 Rounded to Fewer Sig Figs

2 260 2.6 × 102

1 300 3 × 102

5 0

2 years ago

A piece of tape is pulled from a spool and lowered toward a 120-mg scrap of paper. Only when the tape comes within 8.0 mm is the

Anton [14]

Answer:

$1.176\times 10^{-3} N$

Upward

Explanation:

We are given that

Mass of scarp paper, $m=120 mg=120\times 10^{-6}kg$

1mg= $10^{-6} mkg$

Distance,d =8 mm= $8\times 10^{-3} m$

Magnitude of electric force = $F_E= w=mg$

Where $g=9.8 m/s^2$

Substitute the values

$F_E=120\times 10^{-6}\times 9.8$

$F_E=1.176\times 10^{-3} N$

Gravitational force act in downward direction.

The electric force acts in opposite direction and magnitude of electric force is equal to gravitational force.

Hence, the direction of electric force is upward.

6 0

3 years ago

49. A block is pushed across a horizontal surface with a

nata0808 [166]

Answer:

(a) 37.5 kg

(b) 4

Explanation:

Force, F = 150 N

kinetic friction coefficient = 0.15

(a) acceleration, a = 2.53 m/s^2

According to the newton's second law

Net force = mass x acceleration

F - friction force = m a

150 - 0.15 x m g = m a

150 = m (2.53 + 0.15 x 9.8)

m = 37.5 kg

(b) As the block moves with the constant speed so the applied force becomes the friction force.

$F = \mu m g \\\\150 = \mu\times 37.5\\\\\mu = 4$

8 0

3 years ago