In this situation, both of the vehicles turn towards starboard.
By turning to the starboard, or right, side, the vehicles are able to avoid collision. Because waterways are not marked in a manner like roads are, it is necessary to place such conventions of turning in situations where vehicles approach one another head on. If a convention was not in place, the risk of collision would be many times greater. For example, the motorboat operator may turn left, while the PWC operator turns right, resulting in a collision.
The force applied to the second ball by the first ball is 6.734 × 10^-4 N.
<h3>What is impulse of force?</h3>
The impulse of force is defined as the sum of the average force and the duration it is applied.
If the mass of the item remains constant, the impulse of force equals the change in momentum of the object.
Given that: mass of a metal sphere: m = 0.026 kg.
Initial speed of the sphere: u = 3.7 m/s.
When the sphere stops completely, its change in momentum = mu - 0
= 0.026×3.7 N-s.
= 0.0962 N-s.
As the spheres are in contact for 0.007s before the second sphere is shot off down the track, the force applied to the second ball =
change in momentum of 1st ball × time of contact
= 0.0962 × 0.007 N
= 0.0006734 N
= 6.734 × 10^-4 N.
Hence, the force applied to the second ball is 6.734 × 10^-4 N.
Learn more about impulse force here:
brainly.com/question/29787329
#SPJ1
Answer:
The orbital speed of the satellite around the earth in other to remain in perfect circular orbit in given as:
v = sqrt[(Ge*M)/R],
where Ge is the gravitational constant (Ge = 6.673 x 10^-11N/m2/kg2), M is the mass of the earth(m = 5.98 x 10^24kg), and R is the radius of the earth (R = 6.47 x 10^6m)
v = SQRT [ (6.673 x 10^-11 N m2/kg2) • (5.98 x 10^24 kg) / (6.47 x 10^6 m) ]
v = 7.85 x 10^3 m/s
Explanation:
For a satelite in a low altitude orbit around the Earth, the gravitational force is the only force acting of the said satellite keeping it is a circular orbit. To keep this satellite in perfect circular orbit, it must be moving in at a certain speed, which is dependent on the earth mass and radius. This speed can be evaluated from the expression of centripetal force(F = mv2/r). The centripetal force Fc on the satellite is equal to the gravitational force on the satellite from the earth(Fe). That is, (Ge*M*m)/R2 = (m*v2)/R, where M is mass of the earth, and m is the mass of the satellite. making v the subject of the formula, the equation become v = sqrt[(Ge*M)/R].
Answer:
a) Fg = 9.495x10⁻⁶N
b) Fg = 3.908x10⁻⁶N
c)
Explanation:
Given:
m₁ = mass = 3x10⁴kg
r = radius = 1 m
m₂ = 9.3 kg
Questions:
a) What is the magnitude of the gravitational force due to the sphere located at R = 1.4 m, Fg = ?
b) What is the magnitude of the gravitational force due to the sphere located at R= 0.21 m, Fg = ?
c) Write a general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.0 m from the center of the sphere.
a) Since R > r, the equation for the gravitational force is:
Here,
G = gravitational constant = 6.67x10⁻¹¹m³/s² kg
Substituting values:
b) Since R < r, the equation for the gravitational force is:
c) The general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.0 is the same to b)
Answer
given,
I is the loudness of sound
I = 10 Log₁₀ r
r is relative intensity
at when relative intensity is 10⁶
I = 60 dB
how much louder when 100 people would be talking together
I = 10 Log₁₀ r
I = 10 Log₁₀ (10⁶ x 100)
I = 10 Log₁₀ (10⁸)
I = 80 dB
hence, the intensity will be increased by (80 dB -60 dB) 20 dB when 100 people start talking together.
