Answer:
230.4 s
Explanation:
The speed of car A is
and the distance travelled is
so the time taken for car A is
The speed of car B is
and the distance travelled is
so the time taken for car B is
So the difference in time is
Which corresponds to
so car B arrived 230.4 s before car A.
Answer:
V = 381.70 m³
Explanation:
ρ air = 1.28 kg / m³
ρ helium = 0.18 kg / m³
R = 4.5 m
Vb = 0.068 m³
mb = 123 kg
To determine the volume of helium in the balloon when fully inflated
V = 4 / 3 π * R ³
V = 4 * π / 3 ( 4.5 m )³
V = 381.70 m³
To determine the mass total
m = ρ helium * V
m = 0.18 kg / m³ * 381.70 m³
m = 68.70 kg
mt = ( 68.70 + 123 )kg
mt = 191.70 kg
The scheme whereby occupants in a pair of shuttles is as follows
use a strong cable with large weight on the end
Then use the orbital naneuvering system(OMS) to set the whole work as spinning about their common center of gravity.
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The potential energy of the object depends on
- the height of the object with respect to some reference points,
- the mass of the object,
- the gravitational field the object is in.
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Hope it helps ~
<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
