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Natalka [10]
3 years ago
13

g John is walking along a trail when he comes to the bottom of a steep cliff. Before trying to climb up it, he wonders how high

it is. Jake happens to have a bow and arrow with him, and he knows it is able to launch an arrow with a maximum speed of 60.0 m/s. After shooting the arrow straight up with the bow fully extended, he hears it hit the ground at the top of the cliff 7.00 s after he launches it. How high is the cliff
Physics
1 answer:
s344n2d4d5 [400]3 years ago
4 0

Answer:

179.655m

Explanation:

Given

Maximum speed of the arrow v = 60m/s

Time taken to hit the top of the cliff t = 7.0s

Required

Height of the cliff H

Using the equation of motion

H = vt + 1/2gt²

Substitute into the formula:

H = 60(7) + 1/2 (-9.81)(7²) (g is negative due to upward motion of the arrow)

H = 420-4.905(49)

H = 420-240.345

H = 179.655m

Hence the cliff is 179.655m high

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Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
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Answer: 3.7 \times 10^{-4} N

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closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

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Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

4 0
3 years ago
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1000 + 30x30 = 1900. Hope that helps
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Practical steam engines utilize 450ºC steam, which is later exhausted at 270ºC.
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(a) 0.249 (24.9 %)

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The maximum efficiency of the second engine is

\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221

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