(a) It takes approximately 2.8956 seconds for the camera to reach the ground.
(b) The velocity of the camera just before it lands is approximately -28.375 m/s (downward).
We have,
Given information:
Initialheight (h₀) = 41 m (above the ground)
Descentrate of the hot-air balloon = -2.3 m/s (negative because it's descending)
We can use the kinematicequations to solve for the time it takes for the camera to reach the ground and its velocity just before landing.
(a)
To find the time it takes for the camera to reach the ground, we can use the following kinematic equation:
h = h₀ + (v₀)t + (1/2)at²
Where:
h = final height (0 m, as it reaches the ground)
h₀ = initial height (41 m)
v₀ = initial velocity (0 m/s, as the camera is dropped)
a = acceleration (acceleration due to gravity, approximately -9.8 m/s²)
t = time (what we're solving for)
Plugging in the values:
0 = 41 + (0)t + (1/2)(-9.8)t²
Simplifying:
-4.9t² = 41
Divide by -4.9:
t² = -41 / -4.9
t² = 8.36734694
Taking the square root:
t = √8.36734694
t ≈ 2.8956 seconds
(b)
To find the velocity just before the camera lands, we can use the following kinematic equation:
v = v₀ + at
Where:
v = final velocity (what we're solving for)
v₀ = initial velocity (0 m/s)
a = acceleration (acceleration due to gravity, -9.8 m/s²)
t = time (2.8956 seconds, calculated in part a)
Plugging in the values:
v = 0 + (-9.8)(2.8956)
v ≈ -28.375 m/s
The negativesign indicates that the velocity is directed downward, which is consistent with the descending motion.
Thus,
(a) It takes approximately 2.8956 seconds for the camera to reach the ground.
(b) The velocity of the camera just before it lands is approximately -28.375 m/s (downward).
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Answer:
150 hope this helps
Explanation:
Answer:
150
Explanation:
B. Sprouted bean seeds increase.
C. Sprouted bean seeds remain constant.
D. None of the above
As we go up the y-axis, the number of sproutedbean seeds increase (option B).
Graph is a data chart intended to illustrate the relationship between a set (or sets) of numbers (quantities, measurements or indicative numbers) and a reference set.
In a graph, there are two axes as follows;
According to this question, a graph of number of sprouted bean seeds on the y-axis is plotted against temperature on the x-axis.
We can observe that as we go up the y-axis, the number of sprouted bean seeds increase.
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Answer:
The object's initial temperature is 333.6 K
Explanation:
We first assume that the liquid can only transfer heat to the object through convective heat transfer method.
Let T₀ = the initial temperature of the object
T = temperature of the object at anytime.
The rate of heat transfer from the liquid to the object is given as
Q = -hA (T∞ - T)
T∞ = temperature of the fluid = 400 K
A = Surface area of the object in contact with the liquid = 0.015 m²
h = Convective heat transfer coefficient is given to be = 10 W/(m²K)
The rate of heat gained by the object is given by
mC (d/dt)(T∞ - T)
m = mass of the object = ρV
ρ = density of the object = 100 kg/m³
V = volume of the object = 0.000125 m³
m = ρV = 100 × 0.000125 = 0.0125 kg
C = specific heat capacity of the object = 100 J/(kgK)
The rate of heat loss by the liquid = rate of heat gain by the object
-hA (T∞ - T) = mC (d/dt)(T∞ - T)
(d/dt)(T∞ - T) = - (dT/dt) ( Since T∞ is a constant)
- mC (dT/dt) = -hA (T∞ - T)
(dT/dt) = (hA/mC) (T∞ - T)
Let s = (hA/mC)
(dT/dt) = -s (T - T∞)
dT/(T - T∞) = -sdt
Integrating the left hand side from T₀ (the initial temperature of the object) to T and the right hand side from 0 to t
In [(T - T∞)/(T₀ - T∞)] = -st
(T - T∞)/(T₀ - T∞) = e⁻ˢᵗ
(T - T∞) = (T₀ - T∞)e⁻ˢᵗ
s = (hA/mC) = (10 × 0.015)/(0.0125×100) = 0.12
T = 380 K at t = 10 s
T₀ = ?
T∞ = 400 K
st = 0.12 × 10 = 1.2
(380 - 400) = (T₀ - 400) e⁻¹•²
(-20/0.3012) = (T₀ - 400)
(T₀ - 400) = - 66.4
T₀ = 400 - 66.4 = 333.6 K
Hope this Helps!!!
Answer:
option B
Explanation:
we know,
change in energy is equal to
proton mass and the neutron mass are roughly the same
so,
now,
we know,
mass of alpha particle is four times mass of the mass of proton.
mα = 4 m_p
less by a factor of √2
Hence, the correct answer is option B
Answer:
Seismology.
Explanation:
A.
O B.
Total internal reflection is possible when light travels from air to medium B to medium A.
Total internal reflection is possible when light travels from medium A to medium B.
Total internal reflection is possible when light travels from medium B to medium A.
D. Total internal reflection is possible when light travels from air to medium A.
E. Total internal reflection is possible when light travels from air to medium B.
C
C.
If the refractive index of medium A is greater than that of medium B, then total internal reflection is possible when light travels from medium B to medium A.
If the refractive index of medium A is greater than that of medium B, then total internal reflection is possible when light travels from medium B to medium A.