Answer:
Explanation:
According to the newton's second law of motion we can apply F=ma hear
Force = mass * acceleration
(assume the piano is moving left side )
←F = ma
Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of th
ese seeds can be determined with a high-speed camera. In an experiment on one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.6 m/s. The launch angle is measured from the horizontal, with + 90° corresponding to an initial velocity straight up and – 90° straight down.
If a seed is launched at an angle of 0° with the maximum initial speed, how far from the plant will it land? Ignore air resistance, and assume that the ground is flat. (a) 20 cm; (b) 93 cm; (c) 2.2 m; (d) 4.6 m.
If a seed is launched at an angle of 0° with the maximum initial speed, how far from the plant will it land? Ignore air resistance, and assume that the ground is flat. (a) 20 cm; (b) 93 cm; (c) 2.2 m; (d) 4.6 m.
2 answers:
Answer:
Option B, 93 cm
Explanation:
An diagram of the seed's motion is attached to this solution.
This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.
And this would be obtained from the equations of motion,
First of, the height of the plant is related to some quantities of the motion with this relation.
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)
(substituting the t = √(2H/g) derived from above
R = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2
R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.
QED!
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Answer:
-13.18°C
Explanation:
To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.
Its definition is given by the function
Where,
Q = The amount of heat transferred
t = time
k = Thermal conductivity constant
A = Cross-sectional area
The difference in temperature between one side of the material and the other
d= thickness of the material
The problem says that there is a loss of heat twice that of the initial state, that is
Replacing,
Solvinf for ,
Therefore the temprature at the outside windows furface when the heat lost per second doubles is -13.18°C
Answer:
a) a = g / 3
b) x (3.0) = 14.7 m
c) m (3.0) = 29.4 g
Explanation:
Given:-
- The following differential equation for (x) the distance a rain drop has fallen has the form:
- Where, v = Speed of the raindrop
- Proposed solution to given ODE:
v = a*t
Where, a = acceleration of raindrop
Find:-
(a) Using the proposed solution for v find the acceleration a.
(b) Find the distance the raindrop has fallen in t = 3.00 s.
(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.
Solution:-
- We know that acceleration (a) is the first derivative of velocity (v):
a = dv / dt ... Eq 1
- Similarly, we know that velocity (v) is the first derivative of displacement (x):
v = dx / dt , v = a*t ... proposed solution (Eq 2)
v .dt = dx = a*t . dt
- integrate both sides:
∫a*t . dt = ∫dt
x = 0.5*a*t^2 ... Eq 3
- Substitute Eq1 , 2 , 3 into the given ODE:
0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2
= 1.5 a^2 t^2
a = g / 3
- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:
x (t) = 0.5*a*t^2
x (t = 3.0) = 0.5*9.81*3^2 / 3
x (3.0) = 14.7 m
- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:
m (t) = k*x (t)
m (3.0) = 2.00*x(3.0)
m (3.0) = 2.00*14.7
m (3.0) = 29.4 g