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rusak2 [61]
3 years ago
7

What is the total mass of chlorine

Physics
1 answer:
tamaranim1 [39]3 years ago
4 0
One mole of chlorine is equal to 35.453 grams, while one atom is 35.453 AMU
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A soap bubble has a diameter of 4mm .Cakculate the pressure inside it if the atmospheric pressure outside is 10^5Nm^2 . The surf
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Look it up it will give the right anwser
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3 years ago
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A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37 above horizontal]. It gets blocked just after re
stepan [7]
Refer to the diagram shown below.

Neglect air resistance.
The horizontal component of the launch velocity is
 (20 m/s)*cos(37°) = 15.973 m/s
The vertical component of the launch velocity is
 (20 m/s)*sin(37°) = 12.036 m/s

The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by 
(12.036 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (3 m)
12.036t - 4.9t² - 3 = 0
2.4543t - t² - 0.6122 = 0
t² - 2.4563t + 0.6122 = 0
Solve with the quadratic formula.
t = (1/2)[2.4563 +/- √(6.0334 - 2.4488)]
t = 2.1748 or 0.2815 s
The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.

Part a.
The vertical velocity when t = 0.2815 s is
Vy  = 12.036 - 9.8*0.2815
   = 9.2773 m/s
The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is 
√(9.2773² + 15.973² ) = 18.47 m/s
Answer:
The velocity at a height of 3.0 m  is 18.5 m/s (nearest tenth)

Part b.
The horizontal distance traveled is 
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)

6 0
3 years ago
Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m
Kisachek [45]
<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>
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6 0
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Kenny wants to get to Washington DC within 4 hours. Washington DC is 133 miles away from where he is. What is the avg speed he m
umka21 [38]
He must travel 35 mph
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3 years ago
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