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3 years ago

13

Velocity has two pieces of information.what are they

1 answer:

seraphim [82]3 years ago

3 0

They are speed and direction.

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How much heat is needed to melt 13.74 kg of silver that is initially at 20°C?

salantis [7]

Answer:

$Q = 4.63 \times 10^6 J$

Explanation:

As we know that melting point of silver is

T = 961.8 degree C

Latent heat of fusion of silver is given as

L = 111 kJ/kg

specific heat capacity of silver is given as

$s = 240 J/kg C$

now we will have

$Q = ms\Delta T + mL$

$\Delta T = 961.8 - 20$

$\Delta T = 941.8 degree C$

now from above equation

$Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)$

$Q = 3.1 \times 10^6 + 1.53 \times 10^6$

$Q = 4.63 \times 10^6 J$

3 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

A firecracker breaks up into two pieces , one has a mass of 200 g and files off along the x –axis with a speed of 82.0 m/s and t

Readme [11.4K]

Answer:

A) 21.2 kg.m/s at 39.5 degrees from the x-axis

Explanation:

Mass of the smaller piece = 200g = 200/1000 = 0.2 kg

Mass of the bigger piece = 300g = 300/1000 = 0.3 kg

Velocity of the small piece = 82 m/s

Velocity of the bigger piece = 45 m/s

Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s

Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s

since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems

Resultant momentum² = 16.4² + 13.5² = 451.21

Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis ( tan^-1 (13.5 / 16.4)

5 0

3 years ago

A 10 kg bowling ball sits at the top of a 10 m hill and then slides down its icy hillside.

Nataly_w [17]

We need more informtion

5 0

3 years ago

Read 2 more answers

The volume of an ideal gas is adiabatically reduced from 151 L to 80.6 L. The initial pressure and temperature are 1.50 atm and

Zolol [24]

Answer:

gas is dioatomic

T_f = 330.0 K

$\eta = 7.07 mole$

Explanation:

Part 1

below equation is used to determine the type Gas by determining $\gamma$ value

$\frac{V_{1}}{V_{F}}\gamma=\frac{P_{i}}{P_{f}}$

where V_i and V_f is initial and final volume respectively

and P_i and P_f are initial and final pressure

$\gamma = \frac{ln(P_f/P_i)}{ln(V_i/V_f)}$

$\gamma = \frac{ln(3.61/1.50)}{ln(151/80.6}$

\gamma = 1.38

therefore gas is dioatomic

Part 2

final temperature in adiabatic process is given as

$T_f = T_i*[\frac{v_i}{V_f}](^\gamma-1)$

substituing value to get final temperature

$T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}$

T_f = 330.0 K

Part 3

determine number of moles by using following formula

$\eta =\frac{PV}{RT}$

$\eta =\frac{1.013*10^{5}*0.151}{8.314*260}$

$\eta = 7.07 mole$

4 0

3 years ago