Answer:
(B) False
Explanation:
No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning.
Answer:
the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²
Explanation:
Given the data in the question;
initial velocity; u = 0 m/s
height; h = 2.5 m
we find the velocity of the ball just before it touches the foam.
using the equation of motion;
v² = u² + 2gh
we know that acceleration due gravity g = 9.81 m/s²
so we substitute
v² = ( 0 )² + ( 2 × 9.81 × 2.5 )
v² = 49.05
v = √49.05
v = 7.00357 m/s
Now as the ball touches the foam
final velocity v₀ = 0 m/s
compresses S = 3 cm = 0.03 m
so
v₀² = v² + 2as
we substitute
( 0 )² = 49.05 + 0.06a
0.06a = -49.05
a = -49.05 / 0.06
a = -817.5 m/s²
Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²
Answer:
The net downward force on the tank is
Explanation:
Given that,
Area = 1.60 m²
Suppose the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 meters per second per second. The pressure at the surface of the water will be 150 K Pa , and the depth of the water will be 14.4 m . The pressure of the air in the building outside the tank will be 88.0 K Pa.
We need to calculate the net downward force on the tank
Using formula of formula
Where, P = pressure
g = gravity at mars
h = height
A = area
Put the value into the formula
Hence, The net downward force on the tank is
The net downward force on the tank's flat bottom can be found by calculating the pressure at the bottom of the container.
Since the density is constant, the weight can be calculated using the density:
w = mg = pVg = pAhg.
The pressure at the bottom of the container is therefore equal to atmospheric pressure added to the weight of the fluid divided by the area.
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Answer:
x(t) = 20t + 12.75e⁻¹•⁶ᵗ + 487.5
t = 24.375 s
Explanation:
The force balance on the object is given as
Net force = W - Drag force
ma = W - 10v
a = (dv/dt)
ma = m(dv/dt) = 200 - 10v
W = mg
200 = m×32
m = 6.25 kg
m(dv/dt) = 200 - 10v
6.25(dv/dt) = 200 - 10v
(dv/dt) = 32 - 1.6v
v' + 1.6v = 32
Solving this differential equation using the integrating factor method
(ve¹•⁶ᵗ) = ∫ (32e¹•⁶ᵗ) dt
ve¹•⁶ᵗ = (20e¹•⁶ᵗ) + c (where c = constant of integration)
v = (20 + ce⁻¹•⁶ᵗ)
At t = 0, v = 0
0 = 20 + c
c = -20
v = (20 - 20e⁻¹•⁶ᵗ)
v = (dx/dt)
(dx/dt) = 20 - 20e⁻¹•⁶ᵗ
dx = (20 - 20e⁻¹•⁶ᵗ) dt
x(t) = 20t + 12.5e⁻¹•⁶ᵗ + c (c is still the constant of integration)
At t = 0, x = - 500
- 500 = 0 + 12.5 + c
c = 512.5
x(t) = 20t + 12.75e⁻¹•⁶ᵗ - 487.5
when the object hits the ground, x = 0
0 = 20t + 12.75e⁻¹•⁶ᵗ - 487.5
20t + 12.75e⁻¹•⁶ᵗ = 487.5
Solving by trial and error,
t = 24.375 s
Hope this Helps!!!
B. It gives off only one wavelength of electromagnetic radiation
C. It releases only ultraviolet waves of electromagnetic radiation
D. It becomes hotter but gives off less electromagnetic radiation
The black body radiator as it increases in temperature gives off a range of electromagnetic radiation of shorter wavelengths so, the option A is correct.
Radiation is the movement of atomic and subatomic particles as well as waves, such as those that define X-rays, heat rays, and light rays. Radiation of both types, from cosmic and earthly sources, is constantly being thrown at all matter.
The characteristics and behavior of radiation, as well as the matter it interacts with, are outlined in this article, which also explains how energy is transferred from radiation to its surroundings.
The effects of such an energy transfer to living matter, including the typical effects on numerous biological processes, are given a great deal of attention (e.g., photosynthesis in plants and vision in animals).
Thus, the black body radiator gives off a range of electromagnetic radiation of shorter wavelengths.
To know more about radiation:
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Answer: A
Explanation:
Answer is a hope this helps guys!
Stress built up in a rock fault
Heat given off by a forest fire
Water flowing through a hose
Answer:
B
Explanation:
stress built up on a rock fault