Answer:
Explanation:
Acceleration is given by
where
is the change in velocity
is the time interval in which the change in velocity occurs
To find the acceleration at 1 second, we can take the data at t = 1 s and t = 2. We find:
So, the acceleration is
Answer:
I'm just in jss2 but I read physics. this is what I think
Answer:
8.0 N
Explanation:
Force: This can be defined as the mass of a body and its acceleration. The S.I unit of Force is Newton (N).
Mathematically, Fore is expressed as
F = ma ........................... equation 1
Where F = force, m = mass, a = acceleration.
and
I = mΔv
Δv = I/m ............................ Equation 2
Where I = impulse, m = mass, Δv = change in velocity
Given: I = 6.0 Newton-seconds, m = 0.1 kilogram.
Substituting into equation 2
Δv = 6.0/0.1
Δv = 60 m/s.
But
a = Δv/t
where t = time = 0.75 seconds.
a = 60/0.75
a = 80 m/s²
Substitute the values of a and m into equation 1.
F = 0.1(80)
F = 8.0 N.
Thus the average force produced = 8.0 N
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Answer:
t = 3.48 s
Explanation:
The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:
where,
v₀ = initial speed = 110 ft/s
Therefore,
<u>t = 3.48 s</u>
