The answer is option D)
this is because the heat radiated by the flame is mostly absorbed by the air surrounding it, so the air becomes hot and its density decreases (because of expansion), therefore it goes up and it is replaced by cooler air. since all of the hot air flies up, non goes side ways to heat up the match stick, hence it remains cool and does not light up.
option A) also sounds correct, but it isn't. this is because the flame IS hot enough to burn the match stick, it's just that the match stick is positioned the wrong way
Answer:
161.86 N
Explanation:
mass of box m= 55.0 kg
weight of the box, mg= 55×9.81
g here is acceleration due to gravity =9.81 m/sec^2
coefficient of friction between the box and the surface μ= 0.3
the friction force F_s= μmg= 0.3×55×9.81
=161.86 N
to move the ball horizontal force required is 161.86 N
Answer:
Vi = 94.64 m/s
Explanation:
I order to find out the initial velocity of the object, we can use third equation of motion:
2ah = Vf² - Vi²
where,
a = acceleration = -9.8 m/s²
h = maximum height covered by object = 460 m - 3 m = 457 m
Vf = Final Velocity = 0 m/s (since, object momentarily stops at highest point)
Vi = Initial Velocity = ?
Therefore,
2(-9.8 m/s²)(457 m) = (0 m/s)² - Vi²
Vi = √8957.2 m²/s²
<u>Vi = 94.64 m/s</u>
Answer:
In water, the particles are much closer together, and they can quickly transmit vibration energy from one particle to the next.
A water wave is an example of a transverse wave. As water particles move up and down, the water wave itself appears to move to the right or left.
Given Information:
Magnetic field = B = 1×10⁻³ T
Frequency = f = 72.5 Hz
Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m
Required Information:
Maximum Emf = ?
Answer:
Maximum Emf = 20.66×10⁻¹² volts
Explanation:
The maximum emf generated around the perimeter of a cell in a field is given by
Emf = BAωcos(ωt)
Where A is the area, B is the magnetic field and ω is frequency in rad/sec
For maximum emf cos(ωt) = 1
Emf = BAω
Area is given by
A = πr²
A = π(d/2)²
A = π(7.60×10⁻⁶/2)²
A = 45.36×10⁻¹² m²
We know that,
ω = 2πf
ω = 2π(72.5)
ω = 455.53 rad/sec
Finally, the emf is,
Emf = BAω
Emf = 1×10⁻³*45.36×10⁻¹²*455.53
Emf = 20.66×10⁻¹² volts
Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts