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3 years ago

6

A person uses 750 kcal on a long walk calculate the energy used for the walk in kilojoules

1 answer:

guapka [62]3 years ago

6 0

Basically, this problem asks you to convert kilocalories (kcal) to kilojoules (kJ). Both are units of energy. To convert kcal to kJ, the equivalence is: 1 kcal = 4.184 kJ. Through dimensional analysis, the solution is as follows:

750 kcal * 4.184 kJ/1 kcal = 3,138 kJ

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4. Which of the following statement is correct regarding velocity and speed of a moving body?

lara31 [8.8K]

Answer:

Explanation:

Hello friend!!!

The correct option is <u><em>(d) Velocity of a moving body is its speed in a given direction.</em></u>

Hope this helps

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3 years ago

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Answer:

See below

Explanation:

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1 year ago

a stomp rocket takes 1.5 seconds to reach its maximum height what was the initial velocity and what was the maximum height ?

Zinaida [17]

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3 years ago

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the cycle of the moon through its phases, or the synodic month, is a. 21 days long. b. 27 1/3 days long. c. 29 1/2 days long. d.

olasank [31]

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3 years ago

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance

xz_007 [3.2K]

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

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speed of comet(v₁) = 1.6 x 10⁴ m/s

distance (d₁)= 2.7 x 10¹¹ m

to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

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K.E₁ + P.E₁ = K.E₂ + P.E₂

$\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}$

$\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}$

$v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}$

$v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}$

$v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}$

v₂ = 7.6 x 10⁴ m/s

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3 years ago