The speed of the roller coater at the bottom of the hill is 31 m/s.
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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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Answer:
Speed of the satellite V = 6.991 × 10³ m/s
Explanation:
Given:
Force F = 3,000N
Mass of satellite m = 500 kg
Mass of earth M = 5.97 × 10²⁴
Gravitational force G = 6.67 × 10⁻¹¹
Find:
Speed of the satellite.
Computation:
Radius r = √[GMm / F]
Radius r = √[(6.67 × 10⁻¹¹ )(5.97 × 10²⁴)(500) / (3,000)
Radius r = 8.146 × 10⁶ m
Speed of the satellite V = √rF / m
Speed of the satellite V = √(8.146 × 10⁶)(3,000) / 500
Speed of the satellite V = 6.991 × 10³ m/s
Answer:
the observed frequency will reduce but the wavelength will increase
Explanation:
As we know
fo = fs (v/(v-vs))
fo = observed frequency
vs = velocity of source
As per this equation,
When an observer moves away from the stationary source, the observed frequency reduces. Since the observer in the balloon is moving away from the source which itself is moving in opposite direction, the observed frequency will reduce.
Since wavelength = V/fs . The source frequency is unchanged but the velocity is increasing as it is moving in downward direction. Hence, the wavelength will increase
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