Answer:
202.8m
Explanation:
Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.
First calculate the total time travelled by using the second equation of motion
h = Ut + 1/2gt^2
Let assume that u = 0
And h = 3.5
Substitute all the parameters into the formula
3.5 = 1/2 × 9.8 × t^2
3.5 = 4.9t^2
t^2 = 3.5/4.9
t^2 = 0.7
t = 0.845s
To know how far the cannonball travel, let's use the equation
S = UT + 1/2at^2
But acceleration a = 0
T = 2t
T = 1.69s
S = 120 × 1.69
S = 202.834 m
Therefore, the distance travelled by the cannon ball is approximately 202.8m.
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
==>
Givens
=====
V
= 4.00 L
T
= 273oK We're assuming the temperature does not change, just the
pressure.
n
= 0.864 moles
R
= 8.314 joules / mole * oK
P
= ?????
Formula
======
PV
= n*R*T
P
= n*R*T/V
P
= 0.864 * 8.314 * 273 / 4
P
= 490 kpa
You
have to add 1.6 – 0.864 = 0.736 moles of gas.
We
have to assume that the temperature and pressure remain the same when
we add the 0.736 moles of gas. We are now looking for the volume.
PV
= n*R*T
<span>
V
= 0.736 * 8.314 * 273 / 490</span>
V
= 3.41 L Remember this is at about 4 atmospheres so we have to
convert to Standard Pressure.
Total
Volume = 3.41 + 4.00 = 4.41
V1
* P1 = V2 * P2
P1
= 490 kPa
P2
= 101 kPa
V1
= 7.41 L
V2
= ????
<span>
<span>
7.41*
490 = V2 * 101
V2
= 7.41 * 490 / 101
V2
= 35.94 L
</span>
</span>
<span>You
had 4 L now you need 31.94 more.</span>
I need the points please don’t be mad
C
A lean body mass that the cholesterol level in the body is low and that the excess fats in the body are getting rid off as much as one can. In this case, the exercises A.) cardiorespiratory fitness such as jogging and <span>C.) muscular strength and endurance such as body building, abs routines are employed. Answer is D. both A and C.</span>
