I believe this is what you have to do:
The force between a mass M and a point mass m is represented by
So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ = G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ = F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
Explanation:
Distance covered by the particle is given by:
Distance (d) = rate (v) × time (t)
Speed of Mary, v₁ = 50 mph
Speed of Jim, v₂ = 60 mph
It is assumed that, Mary and Jim leave at the same time. After one hour, Jim is 10 miles ahead.
Distance travelled by Jim, d₁ = (60t + 10)
Distance travelled by Mary, d₂ = 50t
The distance between Mary and Jim is greater than or equal to 100 miles.
So, Jim takes is 9 hours more than Mary to cover same distance. Hence, this is the required solution.
Use the concept of beat frequency to find the applicable final freqeuncy for 20Hz beat frequency.
Beat can be defined as 'the interference pattern between two sounds of slightly different frequencies0
The expression for beat frequency is given as
Where,
Final frequency
Initial frequency
The beat frequency for us is 25Hz and the initial frequency is 240Hz, then
Being an absolute value, two values are possible, both in addition and subtraction:
The two possible values are
Answer:
a) The maximum height reached by the projectile is 558 m.
b) The projectile was 21.3 s in the air.
Explanation:
The position and velocity of the projectile at any time "t" is given by the following vectors:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v = (v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time "t"
x0 = initial horizontal position
v0 = initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).
v = velocity vector at time t
a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:
vy = v0 · sin α + g · t
0 = 175 m/s · sin 36.7° - 9.80 m/s² · t
- 175 m/s · sin 36.7° / - 9.80 m/s² = t
t = 10.7 s
Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).
Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.
y = y0 + v0 · t · sin α + 1/2 · g · t²
y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²
y = 558 m
The maximum height reached by the projectile is 558 m
b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.
We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²
0 = t (175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t)
0 = 175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t
- 175 m/s · sin 36.7 / -(1/2 · 9.8 m/s²) = t
t = 21.3 s
The projectile was 21.3 s in the air.
