Answer:
La aceleración del camión que parte del reposo y alcanza la velocidad de 40 km/h en 5 segundos es de 2.22 m/s².
Answer:
The magnitude of the net electric field is:
Explanation:
The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).
On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).
Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>
Let's find first E1 and E2.
The electric field equation is given by:
Where:
- k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
- q1 is the first charge
- d1 is the distance from q1 to P
And E2 will be:
Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.
I hope it helps you!
Answer:
26621 km
Explanation:
We are given;
Mass: m = 5.98 x 10^(24) kg
Period; T = 43200 s
Formula for The velocity(v) of the satellite is:
v = 2πR/T
Where R is the radius
Formula for centripetal acceleration is;
a_c = v²/R
Thus; a_c = (2πR/T)²/R = 4π²R/T²
Formula for gravitational acceleration is:
a_g = Gm/R²
Where G is gravitational constant = 6.674 × 10^(-11) m³/kg.s²
Now the centripetal acceleration of the satellite is caused by its gravitational acceleration. Thus;
Centripetal acceleration = gravitational acceleration.
Thus;
4π²R/T² = Gm/R²
Making R the subject gives;
R = ∛(GmT²/4π²)
Plugging in the relevant values;
R = ∛((6.674 × 10^(-11) × 5.98 x 10^(24) × 43200²)/(4 × π²))
R = 26.621 × 10^(6) m
Converting to km, we have;
R = 26621 km
Answer:
3 V
Explanation:
The relationship between capacity (C), charge stored (Q) and potential difference across a capacitor (V) is
(1)
The problem asks us to find the emf of the battery several seconds after the switch has been closed: this means that the capacitor had enough time to fully charge, so the potential difference across the capacitor (V) is equal to the emf of the battery.
Since we have:
We can solve the equation for V, and we find