Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )
We need to be careful here.
The calculation of the gravitational force between two objects
refers to the distance between their centers.
The minimum possible distance between the Earth's and moon's
centers is the sum of their radii (radiuses).
Earth's radius . . . . . 6,360 km = 6.36 x 10⁶ meters
Moon's radius . . . . . 1,738 km = 1.738 x 10⁶ meters
Sum of their radii = 8.098 x 10⁶ meters
Also:
Earth's mass . . . . . 5.972 x 10²⁴ kg
Moon's mass . . . . . 7.348 x 10²² kg
<span>
and now we're ready to go !
Gravitational force =
G M₁ M₂ / R²
= (6.67 x 10⁻¹¹ N-m²/kg²)(</span><span>5.972 x 10²⁴ kg)(7.348 x 10²² kg)/</span>(8.098 x 10⁶ m)²
= (6.67 · 5.972 · 7.348 / 8.098²) · (10²³) Newtons
= (I get ...) 4.463 x 10²³ Newtons
That's almost exactly 10²³ pounds
= 50,153,000,000,000,000,000 tons.
Those are big numbers.
All I can say is: I wouldn't exactly call that "resting" on the surface".
Answer:
<h3>The answer is 3 kg</h3>
Explanation:
The mass of the object can be found by using the formula

f is the force
a is the acceleration
From the question we have

We have the final answer as
<h3>3 kg</h3>
Hope this helps you
Explanation:
It is given that,
Area of nickel wire, 
Resistance of the wire, R = 2.4 ohms
Initial value of magnetic field, 
Final magnetic field, 
Time, t = 1.12 s
Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :






Induced current in the loop of wire is given by :



So, the induced current in the loop of wire over this time is
. Hence, this is the required solution.