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Montano1993 [528]
3 years ago
14

A hydrogen atom has 1 electron. In which level is it?

Physics
1 answer:
kolezko [41]3 years ago
3 0

If it is in its ground state, it will be in the level 1s

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A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will
alukav5142 [94]

Answer:

The change  in momentum is  \Delta p =   kg \cdot m/s      

Explanation:

From the question we are told that  

       The mass of the probe is  m = 170 kg

       The location of the prob at time t = 22.9 s is  A  =

       The  momentum at time  t = 22.9 s is  p = < 51000, -7000, 0> kg m/s

        The net force on the probe is  F =  N

Generally the change in momentum is mathematically represented as

              \Delta p = F * \Delta t

The initial time is   22.6 s

 The final time  is  22.9 s

             Substituting values  

           \Delta p =  * (22.9 - 22.6)

            \Delta p =  * (0.3)  

              \Delta p =   kg \cdot m/s        

 

6 0
3 years ago
A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s.
irakobra [83]

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system (\tau), measured in Newton-meters, is:

\tau = I\cdot \alpha (1)

Where:

I - Moment of inertia, measured in Newton-meter-square seconds.

\alpha - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

\alpha = \frac{\omega - \omega_{o}}{t} (2)

Where:

\omega_{o} - Initial angular speed, measured in radians per second.

\omega - Final angular speed, measured in radians per second.

t - Time, measured in seconds.

If we know that \tau = 3\,N\cdot m, \omega_{o} = 0\,\frac{rad}{s }, \omega = 145.875\,\frac{rad}{s} and t = 4\,s, then the moment of inertia of the motor is:

\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}

\alpha = 36.469\,\frac{rad}{s^{2}}

I = \frac{\tau}{\alpha}

I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }

I = 0.0823\,N\cdot m\cdot s^{2}

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

5 0
3 years ago
a) What is the average useful power output of a person who does6.00×10^6Jof useful work in 8.00 h? (b) Working at this rate, how
NARA [144]

Answer:

208.33 W

141.26626 seconds

Explanation:

E = Energy = 6\times 10^6\ J

t = Time taken = 8 h

m = Mass = 2000 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height of platform = 1.5 m

Power is obtained when we divide energy by time

P=\frac{E}{t}\\\Rightarrow P=\frac{6\times 10^6}{8\times 60\times 60}\\\Rightarrow P=208.33\ W

The average useful power output of the person is 208.33 W

The energy in the next part would be the potential energy

The time taken would be

t=\frac{E}{P}\\\Rightarrow t=\frac{mgh}{208.33}\\\Rightarrow t=\frac{2000\times 9.81\times 1.5}{208.33}\\\Rightarrow t=141.26626\ s

The time taken to lift the load is 141.26626 seconds

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Irina-Kira [14]

Premature Aging

Mark as brainlist

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