Answer:
High speed optical communication technology
To be able to communicate from the space to the earth and from earth to space is one of the most essential features required during space exploration.
Explanation:
Space exploration involves going into the space, beyond the earth's atmosphere. Landing on other planets and studying their details, going into deeper space beyond the planets to discover new cosmic events or structures is all a part of space exploration.
The key to analyse the studies and observations is being able to communicate the data collected, photos taken etc to the launch centers or space centers on earth. The space centers on earth should also be able to communicate with the persons or the satellites in space.
This is made possible using the optical communication technology which involves the use of optical fibers, lasers etc, since high speeds are more efficient during communication
Answer:
The magnitude of the force required to move the electron through the given field is 2.203 N
Explanation:
Given;
The field strength of the electron, E = 1.375 x 10¹⁹ N/C
charge of electron, q = 1.602 x 10⁻¹⁹ C
The magnitude of the force required to move the electron through the given field is calculated as follows;
F = Eq
F = (1.375 x 10¹⁹ N/C) (1.602 x 10⁻¹⁹ C)
F = 2.203 N
Therefore, the magnitude of the force required to move the electron through the given field is 2.203 N
Answer:
R = 5.28 103 km
Explanation:
The definition of density is
ρ = m / V
V = m /ρ
Where m is the mass and V the volume of the body
The volume of a sphere is
V = 4/3 π r³
Let's replace
4/3 π r³ = m / ρ
R =∛ ¾ m / ρ π
The mass of the planet is
M = 5.5 Me
R = ∛ ¾ 5.5 Me /ρ π
Let's reduce the density to SI units
ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³
ρ = 1.76 10³ kg / m³
Let's calculate
R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)
R = ∛ 0.14723 10²¹
R = 0.528 10⁷ m
R = 0.528 104 km
R = 5.28 103 km
Answer:
Explanation:
Check attachment for solution
Explanation:
It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,
mg is the weight of the car
r is the radius of the curve
v is the speed of the car
Case 1.
F = 640 pounds
Weight of the car, W = mg = 2600 pound
Radius of the curve, r = 650 ft
Speed of the car, v = 40 mph
k = 0.1
Case 2.
Radius of the curve, r = 750 ft
Speed of the car, v = 30 mph
F = 312 N
Hence, this is the required solution.
