Acceleration = change in velocity / time
change in velocity = final velocity - initial velocity
because the object is slowing down, the acceleration will be negative.
acceleration = (4-34) / 60
acceleration = -30 / 60
acceleration = -0.5 m/s^2
answer : the train is accelerating at -0.5 m/s^2
The energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ
U = m ΔT
U = Energy
= Specific heat
m = Mass
ΔT = Change in temperature
ρ = Density
V = Volume
ρ = 1000 g / m³ (Dry air )
= 1 J / g K
ΔT = 5 °C
V = 4 * 4 * 3
V = 48 m³
m = ρ V
m = 1000 * 48
m = 48000 g
U = 1 * 48000 * 7
U = 336000 J
U = 336 kJ
Therefore, the energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ
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Answer:
Explanation:
Nuclear Fission -
The process of the splitting of atom , by radioactive decay process or by any collisional impact .
Sun contains many radioactive isotopes like , thorium , uranium , etc .
The nuclear fission process is an intrinsic property of the radioactive nuclides.
Nuclear Fusion -
The process , where two or more than two atomic nuclei combines to form one or more different atomic nuclei and subatomic particles .
The energy released in both the process is almost same ,
But on the surface of Sun , hydrogen is abundantly present , hence , hydrogen fusion is possible , so the dominant source of energy on the sun is Fusion .
Answer:
runway use is 3307.8 feet
Explanation:
given data
velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s
time = 28 seconds
weight = 28000 lbs
to find out
How many feet of runway was used
solution
we will use here first equation of motion for find acceleration
v = u + at ..............1
here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time
put here value in equation 1
72.016 = 0 + a(28)
a = 2.572 m/s²
and
now apply third equation of motion
s = ut + 0.5×a×t² .......................2
here s is distance and u is initial speed and t is time and a is acceleration
put here all value in equation 2
s = 0 + 0.5×2.572×28²
s = 1008.24 m = 3307.8 ft
so runway use is 3307.8 feet