Why do dzongs have broader base?
1 answer:
You might be interested in
Answer:
A) 1.122 m
B) 5.4 hertz
C) 6.06 m/s ( towards negative x axis )
D) 0.22
E) max speed = amplitude * 34
= 0.22 * 34 = 7.48
minimum speed ( speed at rest ) = 0
Explanation:
The travelling wave = D = 0.22sin ( 5.6x + 34t)
this wave is represented in the form : D = Asin ( kx + wt)
A) wavelength
k = 2 / v
v = wavelength
k = 5.6
therefore wavelength ( v ) from the equation = 2 / 5.6 = 1.122 m
B ) frequency
w = 2f
w = 34
f = frequency
therefore f = w / 2 = 34 / 2
= 5.4 hertz
C ) velocity
speed = f v
v = 5.4 * 1.122 = 6.06 m/s ( towards negative x axis )
D) amplitude
A = 0.22
E ) maximum and minimum speeds of particles on the chord
max speed = amplitude * 34
= 0.22 * 34 = 7.48
minimum speed ( speed at rest ) = 0
Answer:
B.)Angular momentum is always conserved
Explanation:
Angular momentum is given by:
where
m is the mass of the object
v is its speed
r is the distance between the object and the centre of its circular trajectory
In absence of external torques, angular momentum is always conserved. That means that for the spinning star, if its radius r decreases (because it shrinks), in order for L (the angular momentum) to be conserved, the speed (v) must increases, therefore the spinning star speeds up.
So, the correct choice is
B.)Angular momentum is always conserved
Answer:
velocity of truck = 7.32 m/s
velocity of truck = 23.32 m/s
Explanation:
mass of truck, M = 1810 kg
initial velocity of truck, U = 16 m/s
mass of car, m = 673 kg
initial velocity of car, u = 0 m/s
Let the final speed of the car is v and the final speed of truck is V.
Use conservation of momentum
Momentum before collision = Momentum after collision
M x U + m x 0 = M x V + m x v
1810 x 16 + 0 = 1810 V + 673 v
28960 = 1810 V + 673 v .... (1)
As the collision is elastic, so the coefficient of restitution is 1.
By using the formula
where, u1 be the initial velocity of truck, u2 be the initial velocity of the car, v1 be the final velocity of truck and v2 be the final velocity of car.
V - v = - U
V - v = - 16
v - V = 16
v = 16 + V
Substitute in equation (1), we get
28960 = 1810 V + 673 (16 + V)
28960 = 1810 V + 10768 + 673 V
18192 = 2483 V
V = 7.32 m/s
v = 16 + 7.32 = 23.32 m/s
Thus, the velocity of truck after collision is 7.32 m/s and the velocity of car after collision is 23.32 m/s.
Answer:
12.88 s
Explanation:
We can solve this problem by considering only the first half the motion, from the moment the ball is projected upward to the moment it reaches its maximum height.
Since this is a uniformly accelerated motion, we can use the following suvat equation:
where:
s is the vertical displacement
v is the final velocity
t is the time
a is the acceleration
Here we have:
v = 0: the vertical velocity is zero when the ball has reached its maximum height, that is
s = 203 m
(acceleration of gravity, negative because it is downward)
Solving the equation for t, we find the time at which the ball reaches the maximum height:
The second half of the motion is symmetrical and identical to the first one, so the total time in air is just twice this time, therefore: