7. The surface temperature of Venus is 730K, the gravitational constant is 9.8m/s2, and the main...
At the surface of Venus the average temperature is a balmy 460 ∘C due to the greenhouse effect (global warming!), the pressure is 92 Earth-atmospheres, and the acceleration due to gravity is 0.894 gEarth. The atmosphere is nearly all CO2 (molar mass 44.0g/mol) and the temperature remains remarkably constant. We shall assume that the temperature does not change at all with altitude. Part A What is the atmospheric pressure 2.00 km above the surface of Venus? Express your answer in...
At the surface of Venus the average temperature is a balmy 460 ∘C due to the greenhouse effect (global warming!), the pressure is 92 Earth-atmospheres, and the acceleration due to gravity is 0.894 gEarth. The atmosphere is nearly all CO2 (molar mass 44.0g/mol) and the temperature remains remarkably constant. We shall assume that the temperature does not change at all with altitude. Part A What is the atmospheric pressure 4.00 km above the surface of Venus? Express your answer in...
18.54----At the surface of Venus the average temperature is a balmy 460 ∘C due to the greenhouse effect (global warming!), the pressure is 92 Earth-atmospheres, and the acceleration due to gravity is 0.894 gEarth. The atmosphere is nearly all CO2(molar mass 44.0g/mol) and the temperature remains remarkably constant. We shall assume that the temperature does not change at all with altitude. a) What is the atmospheric pressure 4.00 km above the surface of Venus? Express your answer in Earth-atmospheres. Express...
Problem 3.24 (a) Calculate the magnitude of the gravitational force exerted by Venus on a 84 kg human standing on the surface of Venus. (The mass of Venus is 4.9 x 1024 kg and its radius is 6.1 x 106 (b) Calculate the magnitude of the gravitational force exerted by the human on Venus. (c) For comparison, calculate the approximate magnitude of the gravitational force of this human on a similar human who is standing 2.0 meters away (d) What...
1. If gravity on Earth provides a constant acceleration of 9.8m/s2 why aren't we all moving faster and faster toward the center of the Earth all the time? 2. If Magnus Magnuson is in a tug of war match, and he is pulling on the rope with a force of 1265 N, with how much force is the rope pulling on him? 3. Describe the type of force applied by a flat road to a tire when a car is...
Could Venus have ever had liquid oceans? Calculate its surface temperature (w/o greenhouse) using the fact that the Sun was 30% dimmer 4 by ago and Venus is at about 70% the Earth’s distance to the Sun. How does that result compare to the Earth at the present time? Explain your reasoning.
(a) Explain the relationship between the universal gravitational constant G and the acceleration due to gravity at the earth's surface g. Therefore, calculate g from G using the relationships given below. Justify the choice of units for G (Nm kg?). F= mg The mass of the earth is 5.98 x 1024 kg, it's radius is Tg = 6.38 x 10 m, and G = 6.67 x 10-11 Nm?kg (10 marks) (b) Explain, including performing the integral, how the work done...
5 and 7 Question 5 3 ptgu HC Calculate the blackbody surface temperature of Mars (3 sig figs, answer in °C). Solar Flux on Mars (Fs) = 590 W m2 Stefan-Boltzmann constant (0) = 5.67 x 108W m2 K4 Earth's radius = 6.4 x 10 m Assume Mars' albedo = Earth's Albedo (It is worth thinking about whether this is a good assumption, or not! Question 6 O pts Upload a picture of your work from question 5. This is...
1. A car with constant acceleration of 2.3 m/s2 drives toward east (positive x-axces). (a)How much time in seconds is required for the car to change its speed from 10 m/s to 35 m/s? (b)Calculate the total distance in meters, covered during that time (c) Calculate the average speed during that time interval (d) Find x- and y- components of the vector of acceleration 2. From the top of a tower, a person drops a pebble. The peble strikes the...
279 WORKED AND GUIDED PROBLEMS pendulum's period on Earth: 9.8m/S -25. The period of a 0.50-m lunar pendulum is therefore EVALUATE RESULT The longer period on the Moon n accelerations to sense because, as we found in part a, the gravitational accele TE gM 0.50 rn 1.6 m 3.5 s. = (b) We can use the general expression ixth that on Earth. get the desired ratio of periods without having to calculate the on the Moon, 1.6 m/s', is about...