I think speed is the answer
It must have a medium. It must travel in empty space. Mechanical waves are waves which needs medium of propogation.
Given:
Water, 2 kilograms
T1 = 20 degrees Celsius, T2 = 100
degrees Celsius.
Required:
Heat produced
Solution:
Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2 kilograms (4.184 kiloJoules
per kilogram Celsius) (100 degrees Celsius – 20 degrees Celsius)
<u>Q (heat) = 669.42 Joules
</u>This is the amount of heat
produced in boiling 2 kg of water.
The hawk’s centripetal acceleration is 2.23 m/s²
The magnitude of the acceleration under new conditions is 2.316 m/s²
radius of the horizontal arc = 10.3 m
the initial constant speed = 4.8 m/s
we know that the centripetal acceleration is given by
= 
= 23.04/10.3
= 2.23 m/s²
It continues to fly but now with some tangential acceleration
= 0.63 m/s²
therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration
so
= 
= 
= 2.316 m/s²
So the magnitude of net acceleration will become 2.316 m/s².
learn more about acceleration here :
brainly.com/question/11560829
#SPJ4
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
