Answer:
v=53.3m/s
Explanation:
Ek=1/2mv²
7.81×10⁴=1/2×55.0v²
v= the square root of 7.81×10⁴/0.5×55.0
v=53.3m/s
Answer:
775.48 W
Explanation:
given,
diameter of disk = 0.6 cm
length of the disk = 0.4 m
T₁ = 450 K T₂ = 450 K T₃ = 300 K
= 1.33
now,
the value of view factor (F₁₂)corresponding to 1.33
F₁₂ = 0.265
F₁₃ = 1 - 0.265 = 0.735
now,
net rate of radiation heat transfer from the disk to the environment:

= 2 F₁₃ A₁ σ (T₁⁴ - T₃⁴)
= 2 x 0.735 x π x (0.3)² x (5.67 x 10⁻⁸ W/m²) (450⁴ - 300⁴)
= 775.48 W
Net radiation heat transfer from the disks to the environment = 775.48 W
<u>Answer:</u> The mass of the second car is 12666.7 kg
<u>Explanation:</u>
To calculate the mass of car, we use the equation of law of conservation of momentum, which is:

where,
= mass of car 1 = 9500 kg
= Initial velocity of car 1 = 14 m/s
= mass of car 2 = ? kg
= Initial velocity of car 2 = 0 m/s
= Final velocity = 6.0 m/s
Putting values in above equation, we get:

Hence, the mass of the second car is 12666.7 kg
Answer:
0.05 m
Explanation:
From the question given above, the following data were obtained:
Mass of first object (M1) = 9900 kg
Gravitational force (F) = 12 N
Mass of second object (M2) = 52000 kg
Distance apart (r) =?
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Thus, we can obtain the distance between the two objects as shown below:
F = GM1M2/r²
12 = 6.67×10¯¹¹ × 9900 × 52000 /r²
Cross multiply
12 × r² = 6.67×10¯¹¹ × 9900 × 52000
Divide both side by 12
r² = (6.67×10¯¹¹ × 9900 × 52000)/12
Take the square root of both side
r = √[(6.67×10¯¹¹ × 9900 × 52000)/12]
r = 0.05 m
Therefore, the distance between the two objects is 0.05 m