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Llana
1 month ago
9

Assume you have a rocket in Earth orbit and want to go to Mars. The required change in velocity is ΔV≈9.6km/s . There are two op

tions for the propulsion system --- chemical and electric --- each with a different specific impulse. Recall that the relationship between specific impulse and exhaust velocity is: Vex=g0Isp Using the Ideal Rocket Equation and setting g0=9.81m/s2 , calculate the propellant fraction required to achieve the necessary ΔV for each of propulsion system. Part 1: Cryogenic Chemical Propulsion First, consider a cryogenic chemical propulsion system with Isp≈450s . Enter the required propellant fraction as a proportion with at least 2 decimal places (i.e., enter 0.25 to represent 25%): incorrect Part 2: Electric Propulsion Next, consider an electric propulsion system with Isp≈2000s . Enter the required propellant fraction as a proportion with at least 2 decimal places (i.e., enter 0.25 to represent 25%):
Physics
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A champion athlete can produce one horsepower (746 W) for a short period of time. The number of 16-cm-high steps a 70-kg athlete
Sav [3153]

Answer:

407 steps

Explanation:

Based on the question,

P = mgh/t........... Equation 1

Where P stands for power, m is mass, g denotes gravity, h is height, and t represents time.

Rearranging the equation to solve for h, we have:

h = Pt/mg............. Equation 2

Providing values: P = 746 W, t = 1 minute = 60 seconds, m = 70 kg.

Given constant: g = 9.8 m/s²

By substituting into equation 2

h = 746(60)/(70×9.8)

h = 44760/686

h = 65.25 m

h = 6525 cm

Calculating number of steps: 6525/16

The resulting number of steps = 407 steps

6 0
4 months ago
A student solving a physics problem for the range of a projectile has obtained the expression r= v20sin(2θ)g where v0=37.2meter/
ValentinkaMS [3465]

The formula for range is:

R = \frac{v_o^2 sin2\theta}{g}

Given values are:

v_0=37.2m/s

where θ equals 14.1 degrees

g=9.80m/s^2

Using the equation above,

R = \frac{37.2^2 sin2*14.1}{9.80}

The calculated range is 66.7 meters.

Therefore, the range is approximately 66.1 meters.

5 0
4 months ago
Read 2 more answers
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