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4 years ago

By how much does the pressure of a gas in a rigid vessel decrease when the temperature drops from 0 ∘C to –1∘C?

1 answer:

7 0

When a problem says a rigid vessel, it means that volume is constant. At constant V, pressure and temperature are indirectly proportional. We calculate as follows:

P1/T1 = P2/T2
P1/P2 = T1/T2
P1/P2 = 273.15 / 272.15
P1/P2 = 1.00

Hope this helps. Have a nice day.

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A golf club with 65J of kinetic energy strikes a stationary golf ball with a mass of 46g. The energy transfer is only 20% effici

umka21 [38]

Kinetic energy of golf club = 65J,
kinetic energy supplied to golf ball = 20% of 65 = 0.2 * 65 = 13J,
kinetic energy of ball = [mass * Velocity²]/2,
mass = 46gm = 0.046Kg,
[0.046 * V²]/2 = 13, or 0.046 *V² = 26,
V² = 26/0.046 = 565.22,
V = 23.77 m/sec = initial velocity of golf ball after hitting.

4 0

3 years ago

A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at

Readme [11.4K]

Answer:

KE = 1.75 J

Explanation:

given,

mass of ball, m₁ = 300 g = 0.3 Kg

mass of ball 2, m₂ = 600 g = 0.6 Kg

length of the rod = 40 cm = 0.4 m

Angular speed = 100 rpm= $100\times \dfrac{2\pi}{60}$

=10.47\ rad/s

now, finding the position of center of mass of the system

r₁ + r₂ = 0.4 m.....(1)

equating momentum about center of mass

m₁r₁ = m₂ r₂

0.3 x r₁ = 0.6 r₂

r₁ = 2 r₂

Putting value in equation 1

2 r₂ + r₂ = 0.4

r₂ = 0.4/3

r₁ = 0.8/3

now, calculation of rotational energy

$KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2$

$KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)$

$KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)$

$KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)$

KE = 1.75 J

the rotational kinetic energy is equal to 1.75 J

7 0

4 years ago

A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen

NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

$F_1 + F_2 = W$

$F_1 + F_2 = (1530\times 9.81)$

$F_1 + F_2 = 15009.3 N$

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

$F_1(1.15) = F_2(2.70 - 1.15)$

$F_1 = 1.35 F_2$

now from above equation

$F_2 + 1.35F_2 = 15009.3$

now we have

$F_2 = 6392.85 N$

now the other force is given as

$F_1 = 8616.45 N$

4 0

4 years ago

An ax is an example of a ____________.

natta225 [31]

<span>An ax is an example of a wedge. The correct option among all the options that are given in the question is the second option or option "b". The other choices given in the question are incorrect and can be easily neglected. I hope that this is the answer that has actually come to your great help.</span>

4 0

3 years ago

The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2

Darya [45]

Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

d = diameter of copper wire = 2 mm = 2 x 10⁻³ m

Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

$i$ = magnitude of current

magnitude of current is given as

$i = \frac{Ane^{2}\tau E}{m}$

$i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}$

$i$ = 1.87 A

4 0

3 years ago