The new gravitational attraction will be 1/4 as much
Explanation:
The magnitude of the gravitational force between two objects is given by
where
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
In this problem, the original force between the two objects is F, when they are separated by a distance r.
Later, the distance between the two objects is doubled, so the new distance is
Therefore, the new force will be
Therefore, the new force will be one-fourth as much.
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Answer:
26.8 seconds
Explanation:
To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:
v = final velocity
z = initial velocity
x = distance
t = time
a = acceleration
First let's find the final velocity the plane will have at the end of the runway using the first equation:
Now we can plug this into the second equation to find t:
Then using 3 significant figures we round to 26.8 seconds
The period T of a pendulum is given by:
where L is the length of the pendulum while
is the gravitational acceleration.
In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is
. Using this data, we can solve the previous formula to find L:
Answer:
q_poly = 14.55 KJ/kg
Explanation:
Given:
Initial State:
P_i = 550 KPa
T_i = 400 K
Final State:
T_f = 350 K
Constants:
R = 0.189 KJ/kgK
k = 1.289 = c_p / c_v
n = 1.2 (poly-tropic index)
Find:
Determine the heat transfer per kg in the process.
Solution:
-The heat transfer per kg of poly-tropic process is given by the expression:
q_poly = w_poly*(k - n)/(k-1)
- Evaluate w_poly:
w_poly = R*(T_f - T_i)/(1-n)
w_poly = 0.189*(350 - 400)/(1-1.2)
w_poly = 47.25 KJ/kg
-Hence,
q_poly = 47.25*(1.289 - 1.2)/(1.289-1)
q_poly = 14.55 KJ/kg
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is
and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:
