Centripetal acceleration = (speed)² / (radius) .
Force = (mass) · (acceleration)
Centripetal force = (mass) · (speed)² / (radius) .
= (11 kg) · (3.5 m/s)² / (0.6 m)
= (11 kg) · (12.25 m²/s²) / (0.6 m)
= (11 · 12.25) / 0.6 kg-m/s²
= 224.58 newtons. (about 50.5 pounds)
That's the tension in Miguel's arm or leg or whatever part of his body
Jesse is swinging him by. It's the centripetal force that's needed in
order to swing 11 kg in a circle with a radius of 0.6 meter, at 3.5
meters/second. If the force is less than that, then the mass has to
either swing slower or else move out to follow a bigger circle.
Answer:
The pressure of the remaining gas in the tank is 6.4 atm.
Explanation:
Given that,
Temperature T = 13+273=286 K
Pressure = 10.0 atm
We need to calculate the pressure of the remaining gas
Using equation of ideal gas
For a gas
Where, P = pressure
V = volume
T = temperature
Put the value in the equation
....(I)
When the temperature of the gas is increased
Then,
....(II)
Divided equation (I) by equation (II)
Hence, The pressure of the remaining gas in the tank is 6.4 atm.
Lahars can also be formed when high-volume or long-duration rainfall occurs during or after an eruption. It can change the shape of a mountain by blowing parts of it away, but volcanic eruptions can also build up the land around a volcano when lava flows out and hardens on the surface. The surface of the Earth can crack and shift during an earthquake above the point where the crust moves.
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
==>
It would be 3.15 in meters