Answer:
Second drop: 1.04 m
First drop: 1.66 m
Explanation:
Assuming the droplets are not affected by aerodynamic drag.
They are in free fall, affected only by gravity.
I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.
We can use the equation for position under constant acceleration.
X(t) = x0 + v0 * t + 1/2 * a *t^2
x0 = 0
a = 9.81 m/s^2
v0 = 0
Then:
X(t) = 4.9 * t^2
The drop will hit the floor when X(t) = 1.9
1.9 = 4.9 * t^2
t^2 = 1.9 / 4.9
That is the moment when the 4th drop begins falling.
Assuming they fall at constant interval,
Δt = 0.62 / 3 = 0.2 s (approximately)
The second drop will be at:
X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m
And the third at:
X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m
The positions are:
1.9 - 0.86 = 1.04 m
1.9 - 0.24 = 1.66 m
above the floor
Answer:
317.52 mi/hr
Explanation:
First convert Meters into miles as the answer is required in miles/ h
1000m = 0.62 mi
Now, convert second into hours
7.45s = 0.0001 hr
The speed of the boat would be
v = 0.62/0.0001
=317.52 mi/hr
Answer:
0.8J
Explanation:
Given parameters:
Force = 20N
Compression = 0.08m
Unknown:
Spring constant = ?
Elastic potential energy = ?
Solution:
To solve this problem, we use the expression below:
F = k e
F is the force
k is the spring constant
e is the compression
20 = k x 0.08
k = 250N/m
Elastic potential energy;
EPE = 
k e² = x 250 x 0.08²
Elastic potential energy = 0.8J
Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
Answer:
Vi = 5 m/s
Explanation:
let (a) acceleration = 0.75 m/s²
(t) time = 20 seconds
Vf = final velocity = 72 km/hr (convert to m/s to units consistency = 20 m/s)
find Initial velocity (Vi)
Vf - Vi
a = -----------
t
Vi = Vf - (a * t) = 20 - (0.75 * 20)
Vi = 5 m/s
