Answer:
b) 3.72m/s²
c) 9.33*10^5
d) 9.33*10^5
e) 11.85 hrs
Explanation:
a) to confirm that gEarth is about 98 m/s².
Let's use the formula:
= 9.78 m/s²
=> 9.8m/s²
b) Given:
r = 2106 miles
=3.72 m/s²
c) we use:
d) Let's take the force of gravitybon earth due to satellite as our answer in (c) because the Earth's gravitational force on a GPS satellite and the force of gravity on a GPS satellite on earth are equal and opposite (two mutual forces).
e) In a circular motion,
Gravitional force = Centripetal force.
Solving for v, we have
v = 3886m/s
Therefore,
v = 2πR/T
Solving for T, we have:
T = 42650seconds
Convert T to hours
T = 42650/60*60
T = 11.86hrs
Answer:
7212.3 N
Explanation:
F = 7.38 x 10^4 N, v = 36.2 m/s
Let b be the strength of magnetic field and charge on the particle is q.
F = q v B Sin theta
Here theta = 90 degree
7.38 x 10^4 = q x 36.2 x B x Sin 90
q B = 2038.7 .....(1)
Now, theta = 17 degree, v = 12.1 m/s
F = q v B Sin theta
F = 2038.7 x 12.1 x Sin 17 ( q v = 2038.7 from equation (1)
F = 7212.3 N
0.520155077123917 i believe?? hope this helps
Answer:
Thats the third law______
<em>Answer: D</em>
<em>Given data:</em>
The electric field due to charge q₁ is (E₁) = 1.5 × 10⁵ N/C
The electric field due to charge q₂ is (E₂) = 7.2 × 10⁵ N/C
Determine the net electric field at point P (Enet) = ?
electric field measured in Newtons/coulomb
<em>We know that,</em>
The net electric field (Enet) is vector sum of E₁ and E₂ caused by the charge q₁ and q₂ respectively
Enet = E₁ + E₂
= (1.5 × 10⁵)+ (7.2 × 10⁵)
= 8.7 × 10⁵ Newtons/coulomb
<em>The net electric field at point P is 8.7 × 10⁵ N/C</em>
