Air at 27 ºC and 5 atm is throttled by a valve to 1 atm. If the valve is adiabatic and the change in kinetic energy is negligibl
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Answer:
-5.1 kg m/s
Explanation:
Impulse is the change in momentum.
Change in momentum= final momentum - initial momentum=m +m
Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)
Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)
Answer:
q₁= +0.5nC
Explanation:
Theory of electrical forces
Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
To solve this problem we apply Coulomb's law:
Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
o solve this problem we apply Coulomb's law:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁,q₂:Charges in Coulombs (C)
d: distance between the charges in meters
Data:
Equivalences
1nC= 10⁻⁹ C
1cm= 10⁻² m
Data
q₃=+5.00 nC =+5* 10⁻⁹ C
q₂= -2.00 nC =-2* 10⁻⁹ C
d₂= 5.00 cm= 5*10⁻² m
d₁= 2.50 cm= 2.5*10⁻² m
k = 8.99*10⁹ N*m²/C²
Calculation of magnitude and sign of q1
Fn₃=0 : net force on q3 equals zero
F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.
F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.
We propose the algebraic sum of the forces on q₃
F₂₃ - F₁₃=0
We eliminate k*q₃ of the equation
q₁= +0.5*10⁻⁹ C
q₁= +0.5nC
Answer:
d = 0.645 m <em>(assuming a radius of the ball bearing of 3 mm)</em>
Explanation:
<u>The given information is:</u>
- <em>The distance from the center of the sun to the center of the earth is 1.496x10¹¹m =
</em>
- <em>The radius of the sun is 6.96x10⁸m =
</em>
<u>We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm = </u>.
First, we need to find how many times the radius of the sun is bigger respect to the radius of the ball bearing, which is given by the following equation:
Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth, :
[tex] d_{s} = \frac{d_{e}}{r_{s}/r_{b}} = \frac{1.496 \cdot 10^{11} m}{2.32\cdot 10^{11}} = 0.645 m
I hope it helps you!