Answer:
(a) m = 1.6 x 10²¹ kg
(b) K.E = 2.536 x 10¹¹ J
(c) v = 7.12 x 10⁵ m/s
Explanation:
(a)
First we find the volume of the continent:
V = L*W*H
where,
V = Volume of Slab = ?
L = Length of Slab = 4450 km = 4.45 x 10⁶ m
W = Width of Slab = 4450 km = 4.45 x 10⁶ m
H = Height of Slab = 31 km = 3.1 x 10⁴ m
Therefore,
V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)
V = 6.138 x 10¹⁷ m³
Now, we find the mass:
m = density*V
m = (2620 kg/m³)(6.138 x 10¹⁷ m³)
<u>m = 1.6 x 10²¹ kg</u>
<u></u>
(b)
The kinetic energy will be:
K.E = (1/2)mv²
where,
v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)
v = 3.17 x 10⁻¹⁰ m/s
Therefore,
K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²
<u>K.E = 2.536 x 10¹¹ J</u>
<u></u>
(c)
For the same kinetic energy but mass = 77 kg:
K.E = (1/2)mv²
2.536 x 10¹¹ J = (1/2)(77 kg)v²
v = √(2)(2.536 x 10¹¹ J)
<u>v = 7.12 x 10⁵ m/s</u>
The current in the 50 Ω resistor is A) 1.2 A
The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
Answer:
Sohan told Geeta that i had done my work
