The frequency of the wave will not change. Since the change in medium doesn't affect the source of the waves, the frequency of those waves do not change.
Hope this helps! :)
1 g = 1 ÷ 1000 kg
= 0.001 kg
1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³
1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³
The density is 1000 kg/m³.
<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>
Answer:
1.42
Explanation:
<em> got it right on my homework </em>
Answer:
T = 693.147 minutes
Explanation:
The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.
Therefore, the total salt in the tank at time t = 1000x kg
Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L
Hence, the amount of salt that is added to the tank per minute = 
Also, there is a continuous outflow from the tank at a rate of 6 L/min.
Hence, amount of salt subtracted from the tank per minute = 6x kg/min
Now, the rate of change of salt concentration in the tank = 
So, the rate of change of salt in the tank can be given by the following equation,
or, 
or, T = 693.147 min (time taken for the tank to reach a salt concentration
of 0.05 kg/L)
