<span>After an exoplanet has been identified using a given detection method, scientists attempt to identify the basic properties of the planet which can tell us what it might be made of, how hot it might be, whether or not it contains an atmosphere, how that atmosphere might behave, and finally, whether the planet may be suitable for life. It is often useful to first determine basic properties of the parent star (such as mass and distance from the Earth). This is then followed by the use of planetary detection methods to calculate planetary mass, radius, orbital radius, orbital period, and density. The density calculation will provide clues as to what the planet is made of and whether or not it contains a significant atmosphere.
Mass and Distance of Parent Star
The mass and distance of an exoplanet's parent star must often be calculated first, before certain measurements of the exoplanet can be made. For example, determining the star's distance is an important step in determining a star's mass (see below). Knowing the mass of a star then allows the mass of the planet to be measured, for example when using the Radial Velocity Method.</span>
Answer:
The speed of the combined vehicles is 6.82m/s
Explanation:
Using the law of conservation of momentum which stayed that the sum of momentum of bodies before collision is equal to their sum of momentum after collision. After collision, both object moves with the same velocity.
Momentum = mass×velocity
Before collision:
Momentum of vehicle or mass 3000kg moving with velocity 25m/s
= 3000×25
= 75000kgm/s
Pa = 75000kgm/s
Momentum of vehicle with mass 2500kg moving with velocity of -15m/s
= 2500×-15
= -37500kgm/s
After collision:
Momentum = (3000+2500)V
Where v is their common velocity
Momentum after collision = 5500V
Based on the law:
75000+(-37500) = 5500V
75000-37500 = 5500V
37500 = 5500V
V = 37500/5500
V = 6.82m/s
Answer:
20,000,000 N
Explanation:
First find the acceleration:
a = Δv / Δt
a = (0 − 40 m/s) / 0.010 s
a = -4000 m/s²
Next use Newton's second law to find the force on the car:
F = ma
F = (5000 kg) (-4000 m/s²)
F = -20,000,000 N
According to Newton's third law, the force on the wall is equal and opposite the force on the car.
F = 20,000,000 N
Answer:
d) 700 m/s
Explanation:
if k is the force constant and x is the maximum compression distance, then:
the potential energy the spring can acquire is given by:
U = 1/2×k×(x^2)
and, the kinetic energy system is given by:
K = 1/2×m×(v^2)
if Ki is the initial kinetic energy of the system, Ui is the initial kinetic energy of the system and Kf and Uf are final kinetic and potential energy respectively then, According to energy conservation:
initial energy = final energy
Ki +Ui = Kf +Uf
Ui = 0 J and Kf = 0J
Ki = Uf
1/2×m×(v^2) = 1/2×k×(x^2)
m×(v^2) = k×(x^2)
v^2 = k×(x^2)/m
= (500)×((21×10^-2)^2)/(19×10^-3 + 8)
= 2.75
v = 1.66 m/s
the v is the final velocity of the bullet block system, if m1 is the mass of bullet and M is the mass of the block and v1 is the initial velocity of the bullet while V is the initial velocity of the block, then by conservation linear momentum:
m1×v1 + M×V = v×(m1 + M) but V = 0 because the block is stationary, initially.
m1×v1 = v×(m1 + M)
v1 = v×(m + M)/(m1)
= (1.66)×(19×10^-3 + 8)/(19×10^-3)
= 699.86 m/s
≈ 700 m/s
Therefore, the velocity of the bullet just before it hits the block is 700 m/s.
....possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors
