The speed of the roller coater at the bottom of the hill is 31 m/s.
<h3>
Speed of the roller coater at the bottom of the hill</h3>
Apply the principle of conservation of mechanical energy as follows;
K.E(bottom) = P.E(top)
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
- v is the speed of the coater at bottom hill
- h is the height of the hill
- g is acceleration due to gravity
v = √(2 x 9.8 x 49)
v = 31 m/s
Thus, the speed of the roller coater at the bottom of the hill is 31 m/s.
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Yes, in general, metalloids are slightly reactive.
Internal energy of the system changes by ΔE = 178 J.
Heat given to the system = Q = +658 J.
According to the first law of thermodynamics,
ΔE = Q + W
178 = 658 + W
∴ W = 178-658 = -480 J
Minus sign indicates that work is done by the system.
Answer:
<h3>473.8 m/s; 473.8 m/s</h3>
Explanation:
Given the initial velocity U = 670m/s
Horizontal velocity Ux = Ucos theta
Vertical component of the cannon velocity Uy = Usin theta
Given
U = 670m/s
theta = 45°
horizontal component of the cannonball’s velocity = 670 cos 45
horizontal component of the cannonball’s velocity = 670(0.7071)
horizontal component of the cannonball’s velocity = 473.757m/s
Vertical component of the cannonball’s velocity = 670 sin 45
Vertical component of the cannonball’s velocity = 670 (0.7071)
Vertical component of the cannonball’s velocity = 473.757m/s
Hence pair of answer is 473.8 m/s; 473.8 m/s
Answer:
The acceleration is 2.2 m/s^2
Explanation:
In the attached image, we can see the free body diagram. And using the second law of Newton it will be possible to find the acceleration of the box.