Which of these warnings refers to a chemical property of the material?

Ikg.

Tcecarenko [31]

Answer:

The mass of G1 at all times during this trial was 0.5 kg.

The velocity of G1 + G2 after the collision was -1.24 m/s.

The momentum of G1 after the collision was -2.10 kg · m/s.

Explanation:

i got it right

6 0

2 years ago

Snow tends to sublime outside during cold, clear winter days when the atmospheric pressure is low. During this proces

olga nikolaevna [1]

A I think because sublimation is when something skips melting and goes directly to vapor

3 0

2 years ago

The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

Cerrena [4.2K]

Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

$I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }$

$I = 19.799\,kg\cdot m^{2}$

b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

$\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}$

$\theta_{T} = 344.580\,rad$

The total number of revolutions of the wheel is:

$n_{T} = \frac{\theta_{T}}{2\pi}$

$n_{T} = \frac{344.580\,rad}{2\pi}$

$n_{T} \approx 54.842\,rev$

5 0

2 years ago

How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of ele

konstantin123 [22]

Answer:

-1486 KJ

Explanation:

The work done by an electric field on a charged body is:

W = ΔV * q

where ΔV = change in voltage

q = total charge

The total charge of Avogadro's number of electrons is:

6.0221409 * 10^(23) * -1.6023 * 10^(-19) = -9.65 * 10^(4)

The change in voltage, ΔV, is:

9.20 - (6.90) = 15.4

Therefore, the work done is:

W = -9.65 * 10^(4) * 15.4 = -1.486 * 10^6 J = -1486 KJ

The negative sign means that the motion of the electrons is opposite the electrostatic force.

5 0

2 years ago

The spaceship Intergalactic landed on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist

max2010maxim [7]

Answer:

18808.7 m/s^2

Explanation:

Given

Length of the pendulum L = 1.44 m

Number of complete cycles of oscillation n = 1.10 x 10^2

total time of oscillation t = 2.00 x 10^2 s

The period of the T = n/t

T = (1.10 x 10^2)/(2.00 x 10^2) = 0.55 ^-s

The period of a pendulum is gotten as

T = $2\pi \sqrt{\frac{L}{g} }$

where g is the acceleration due to gravity

substituting values, we have

0.55 = $2\pi \sqrt{\frac{1.44}{g} }$

0.0875 = $\sqrt{\frac{1.44}{g} }$

squaring both sides of the equation, we have

7.656 x 10^-3 = 144/g

g = 144/(7.656 x 10^-3) = 18808.7 m/s^2

3 0

2 years ago