Answer:
<h3>1.03684m</h3>
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g where
R is the distance moves in horizontal direction = 18.4m
H is the height
U is the velocity of the baseball = 40m/s
g is the acceleration due to gravity = 9.8m/s²
Substitute the given parameters into the formula and calculate H as shown;
18.4 = 40√2H/9.8
18.4/40 = √2H/9.8
0.46 = √2H/9.8
square both sides;
(0.46)² = (√2H/9.8)²
0.2116 = 2H/9.8
2H = 9.8*0.2116
2H = 2.07368
H = 2.07368/2
H = 1.03684m
Hence the ball is 1.03684m below the launch height when it reached home plate.
According to Boyle’s law, For a fixed amount of an ideal gas kept at a fixed temperature, P (pressure) and V (volume) are inversely proportional.
Therefore,

Given
,
and
.
Thus,

<span>Even in space, there is still presence of gravity. The
cause of weightlessness is not how far above the earth the space shuttle is but
rather how fast it is travelling. The shuttle is in free fall causing
weightlessness, but it is travelling fast enough to miss the earth as it falls.
Similarly, the airplane could also provide weightlessness if it went free fall
as well. However, that ends as the plane hits the ground. </span>
Explanation:
V=u+at
where,
v=final speed
u=initial speed,(starting speed)
a=acceleration
t=time
- v=u+at = 6=2+a*2
6=2+2a
2a=6-2
2a=4
a=4/2 = 2
a =2
2. to find time taken
v=u+at
25=5*2t
2t=25-5
2t=20
t=20/2
t=10sec
3. finding final speed
v=u+at
v=4+10*2
=4+20
v=24m/sec
5.v=u+at
=5+8*10
=5+80
V=85m/sev
6. v=u+at
8=u+4*2
8=u+8
U=8/8
u=1
these are your missing values
We take the derivative of Ohm's law with respect to time: V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:
-0.03 V/s = (0.04 A)(0.04 ohm/s) + (200 ohms)(dI/dt)
dI/dt = -0.000158 = -1.58 x 10^-4 A/s