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neonofarm [45]
4 years ago
9

The position of the front bumper of a test car under microprocessor control is given by x(t) = 2.31 m + (4.90 m/s2)t2 - (0.100 m

/s6)t6. (a) find its position at the instants when the car has zero veloc
Physics
1 answer:
vovangra [49]4 years ago
3 0
The equation of the car is given by the equation,

                          x(t) = 2.31 + 4.90t² - 0.10t⁶

If we are going to differentiate the equation in terms of x, we get the value for velocity.

                  dx/dt = 9.8t - 0.6t⁵

Calculate for the value of t when dx/dt = 0.

                 dx/dt = 0 = (9.8 - 0.6t⁴)(t)

The values of t from the equation is approximately equal to 0 and 2. 

If we substitute these values to the equation for displacement,

(0)   , x = 2.31 + 4.90(0²) - 0.1(0⁶) = 2.31

(2)    , x = 2.31 + 4.90(2²) - 0.1(2⁶) = 15.51

Thus, the positions at the instants where velocity is zero are 2.31 and 15.51 meters. 
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Using ohm's law, determine the amount of power dissipated in a 5.6k ohm resistor connected to a 5v dc source.
Karolina [17]
The equation for power is P=VI, we can then rearrange V=IR and substitute I into the equation, this gets us V/R=I. Thus we get P=V^2/R

Next we convert kΩ to Ω. This tells us the resistor in Ω is 5600Ω.

Finally, plug into equation, we get P = (5V^2)/5600Ω.

P = 0.0044642857W

Hope this helps!
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3 years ago
To apply the Biot-Savart law to find the magnetic field produced on the z axis from current elements in the xy plane. In this pr
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6 0
3 years ago
When Jenny writes the pencil exerts a force of 5N on the paper 
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3 years ago
Suppose that the dipole moment associated with an iron atom of an iron bar is 2.9 × 10-23 J/T. Assume that all the atoms in the
allsm [11]

Answer:

16.042336 Am²

27.2719712 Nm

Explanation:

Dipole moment association with iron atom

\frac{\mu}{N}=2.9\times 10^{-23}\ J/T

L = Length of bar = 6.5 cm

A = Area of bar = 1 cm²

N_A = Avogadro constant = 6.022\times 10^{23}

\rho = Density of iron = 7.9 g/cm³

M = Molar mass = 55.9 g/mol

B = Magnetic field = 1.7 T

Number of atoms is given by

N=\frac{\rho LA}{M}\times N_A\\\Rightarrow N=\frac{7.9\times 6.5\times 1}{55.9}\times 6.022\times 10^{23}\\\Rightarrow N=5.53184\times 10^{23}\ atoms

Dipole moment is given by

\frac{\mu}{N}\times n\\ =2.9\times 10^{-23}\times 5.53184\times 10^{23}\\ =16.042336\ Am^2

The dipole moment of the bar is 16.042336 Am²

Torque is given by

\tau=\mu Bsin\theta\\\Rightarrow \tau=16.042336\times 1.7sin 90\\\Rightarrow \tau=27.2719712\ Nm

The torque exerted to hold this magnet perpendicular to an external field is 27.2719712 Nm

8 0
3 years ago
Which of the the following quantities has the same unit as kilowatt - hour?
Advocard [28]

Answer:

I think (d) is your answer...

6 0
3 years ago
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