Let m = mass of asteroid y.
Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.
Given:
F = 6.2x10⁸ N
d = 2100 km = 2.1x10⁶ m
Note that
G = 6.67408x10⁻¹¹ m³/(kg-s²)
The gravitational force between the asteroids is
F = (G*m*(m/3))/d² = (Gm²)/(3d²)
or
m² = (3Fd²)/G
= [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))
= 1.229x10³² kg²
m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)
Answer: 1.1x10¹⁶ kg
85 + 45 = 130
130km in 1.5 hours
130 / 1.5 = 86.6666666666666
the average speed was:
86.6666666666666
a brainliest would be appreciated
This collects light, and bends it into focus.
Answer:
FC vector representation
Magnitude of FC
Vector direction FC
degrees: angle that forms FC with the horizontal
Explanation:
Conceptual analysis
Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.
The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.
degrees
To calculate the magnitudes of the forces we apply Coulomb's law:
Equation (1): Magnitude of the electric force of the charge qA over the charge qC
Equation (2)
: Magnitude of the electric force of the charge qB over the charge qC
Known data
Problem development
In the equations (1) and (2) to calculate FAC Y FBC:
Components of the FBC force at x and y:
Components of the resulting force acting on qC:
FC vector representation
Magnitude of FC
Vector direction FC
degrees: angle that forms FC with the horizontal
The planet MARS is visible without a telescope on many clear nights. The planets JUPITER, MERCURY, VENUS and SATURN are also viewable without the aid of magnification.