Answer:
t = 1.41 sec.
Explanation:
If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.
First, we need to find the value of acceleration, which is the same for both blocks.
If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:
F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)
⇒ a = ( = g/5 m/s²
Once we got the value of a, we can use for instance this kinematic equation, and solve for t:
Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.
Explanation:
It is given that,
The period of the carrier wave, T = 0.01 s
Let f and are frequency and the wavelength of the wave respectively. The relationship between the time period and the frequency is given by :
f = 100 Hz
The wavelength of a wave is given by :
So, the frequency and wavelength of the carrier wave are 100 Hz and respectively. Hence, the correct option is (c).
Answer:
the average force exerted on the ball by the bat is 11,613.27 N
Explanation:
Given;
mass of the baseball, m = 151 g = 0.151 kg
initial velocity of the baseball, u = 39.5 m/s
final velocity of the baseball, v = 45.1 m/s
time of action, t = 1.10 ms = 1.10 x 10⁻³ s
The average force exerted on the ball by the bat is calculate as;
Therefore, the average force exerted on the ball by the bat is 11,613.27 N
Forward thrust has positive values and reverse thrust has negative values.
Thrust is a sudden push or pull in a certain direction.
a)
Flight speed u = 150 km/h
1 km/h = km/s
therefore, 150 km/h = 41.67 km / s
The thrust force represents the horizontal or x-component of momentum equation:
T = 50 * (150 - 41.67)
T = 5416.67 N
Therefore, the value of forward thrust is 5416.67 N.
b)
Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component,
thus thrust equation is:
T = 50 * (0 - 41.67)
T = -2083.5 N
Therefore, the thrust force T is -2083.5 N in the reverse direction.
c)
Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero because
T = 0
Therefore, there is no difference in two velocities in x direction.
The given question is incomplete, the complete question is,
"The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse thrust if the jet is properly deflected. Suppose that while the aircraft rolls down the runway at 150 km/h the idling engine consumes air at 50 kg/s and produces an exhaust velocity of 150 m/s.
a. What is the forward thrust of this engine?
b. What are the magnitude and direction (i.e., forward or reverse) if the exhaust is deflected 90 degree without affecting the mass flow?
c. What are the magnitude and direction of the thrust (forward or reverse) after the plane has come to a stop, with 90 degree exhaust deflection and an airflow of 40 kg/s?"
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