This question involves the concepts of Newton's Law of Gravitation and mass.
The force on Procyon A from Procyon B will be "equal" to the force on Procyon B from Procyon A, which has a value of "3.75 x 10²⁶ N".
Applying Newton's Law of Gravitation, we can find the force on Procyon A from Procyon B, which is equal to the force on Procyon B from Procyon A:
![F=\frac{Gm_1m_2}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BGm_1m_2%7D%7Br%5E2%7D)
where,
F = force = ?
G = universal gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg²
m₁ = mass of Procyon A = 3 x 10³⁰ kg
m₂ = mass of Procyon B = (2.5)(3 x 10³⁰ kg) = 7.5 x 10³⁰ kg
r = distance between them = 2 x 10¹² m
Therefore,
![F=\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(3\ x\ 10^{30}\ kg)(7.5\ x\ 10^{30}\ kg)}{(2\ x\ 10^{12}\ m)^2}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B%286.67%5C%20x%5C%2010%5E%7B-11%7D%5C%20N.m%5E2%2Fkg%5E2%29%283%5C%20x%5C%2010%5E%7B30%7D%5C%20kg%29%287.5%5C%20x%5C%2010%5E%7B30%7D%5C%20kg%29%7D%7B%282%5C%20x%5C%2010%5E%7B12%7D%5C%20m%29%5E2%7D)
<u>F = 3.75 x 10²⁶ N</u>
Learn more about Newton's Law of Gravitation here:
brainly.com/question/17931361?referrer=searchResults
Answer:
a) 43.20V
b) 2.71W/s
c) 40.25s
d) 7.77Nm
Explanation:
(a) The emf of a rotating coil with N turns is given by:
![emf=NBA\omega sin(\omega t)](https://tex.z-dn.net/?f=emf%3DNBA%5Comega%20sin%28%5Comega%20t%29)
N: turns
B: magnitude of the magnetic field
A: area
w: angular velocity
the emf max is given by:
![emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V](https://tex.z-dn.net/?f=emf_%7Bmax%7D%3DNBA%5Comega%3D%2850%29%281.80T%29%280.200m%2A0.100m%29%2824.0rad%2Fs%29%5C%5C%5C%5Cemf_%7Bmax%7D%3D43.20V)
(b) the maximum rate of change of the magnetic flux is given by:
![\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5CPhi_B%7D%7Bdt%7D%3D%5Cfrac%7Bd%28A%5Ccdot%20B%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%28ABcos%5Comega%20t%29%3DAB%5Comega%20sin%28%5Comega%20t%29%5C%5C%5C%5C%5Cfrac%7Bd%5CPhi_B%7D%7Bdt%7D_%7Bmax%7D%3D%28%5Cpi%280.200%2A0.100%29%29%281.80T%29%2824.0rad%2Fs%29%3D2.71%5Cfrac%7BW%7D%7Bs%7D)
(c) ![emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V](https://tex.z-dn.net/?f=emf%28t%3D0.050s%29%3D%2850%29%281.80T%29%280.200m%2A0.100m%29%2824rad%2Fs%29sin%2824.0rad%2Fs%280.050s%29%29%5C%5C%5C%5Cemf%28t%3D0.050s%29%3D40.26V)
(d) The torque is given by:
![\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm](https://tex.z-dn.net/?f=%5Ctau%3DNABIsin%5Ctheta%5C%5C%5C%5CNAB%5Comega%3Demf_%7Bmax%7D%5C%5C%5C%5C%5Ctau%3D%5Cfrac%7Bemf_%7Bmax%7D%7D%7B%5Comega%7D%5Cfrac%7Bemf_%7Bmax%7D%7D%7BR%7D%5C%5C%5C%5C%5Ctau%3D%5Cfrac%7B%2843.20V%29%5E2%7D%7B%2824.0rad%2Fs%29%2810.0%5COmega%29%7D%3D7.77Nm)
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Answer:
correct option is d) 7.0 x 10^-7 N
Explanation:
given data
distance = 175 picometers = 1.75 ×
m
to find out
electrical force
solution
we know atomic no of uranium is 92
and charge on electron is = 1.6 ×
C
and electrical force is express as
electrical force =
.............1
put here value we get
electrical force = ![9*10^9 \frac{92*(1.6*10^{-19})^2}{(1.75*10^{-10})^2}](https://tex.z-dn.net/?f=9%2A10%5E9%20%5Cfrac%7B92%2A%281.6%2A10%5E%7B-19%7D%29%5E2%7D%7B%281.75%2A10%5E%7B-10%7D%29%5E2%7D)
electrical force = 6.921 ×
N
so correct option is d) 7.0 x 10^-7 N
(a.) The acceleration of the car can be calculated using the equation,
2ad = v²
Substituting,
2(a)(180 m) = (22.2 m/s)²
From the equation above, we achieve the value of a = 1.369 m/s²
(b) The force necessary is calculated below,
F = (1,300 kg)(1.369 m/s²)
F = 1,779.7 N
(c) Coefficient of friction,
1,779.7 N = (1,300)(9.81)(x)
The value of x from the generated equation,
x = 0.14.