Answer:
The answer is below
Explanation:
a) The vertical displacement = Δy = 21.5 m - 1.5 m = 20 m
The horizontal displacement = Δx = 69 m wide
Using the formula:
![\Delta y = u_yt+ \frac{1}{2}a_yt^2\\ \\u_y=initial\ velocity\ of \ car\ in\ y\ direction = 0,a_y=g=acceleration\ due\ to\ gravity\\=10m/s^2\\\\\Delta y = \frac{1}{2}a_yt^2\\\\\Delta y=\frac{1}{2}a_yt^2\\\\t=\sqrt{\frac{2\Delta y}{a_y} }=\sqrt{\frac{2*20}{10} } =2\ m/s](https://tex.z-dn.net/?f=%5CDelta%20y%20%3D%20u_yt%2B%20%5Cfrac%7B1%7D%7B2%7Da_yt%5E2%5C%5C%20%5C%5Cu_y%3Dinitial%5C%20velocity%5C%20of%20%5C%20car%5C%20in%5C%20y%5C%20direction%20%3D%200%2Ca_y%3Dg%3Dacceleration%5C%20due%5C%20to%5C%20gravity%5C%5C%3D10m%2Fs%5E2%5C%5C%5C%5C%5CDelta%20y%20%3D%20%20%5Cfrac%7B1%7D%7B2%7Da_yt%5E2%5C%5C%5C%5C%5CDelta%20y%3D%5Cfrac%7B1%7D%7B2%7Da_yt%5E2%5C%5C%5C%5Ct%3D%5Csqrt%7B%5Cfrac%7B2%5CDelta%20y%7D%7Ba_y%7D%20%7D%3D%5Csqrt%7B%5Cfrac%7B2%2A20%7D%7B10%7D%20%7D%20%20%3D2%5C%20m%2Fs)
Also:
![\Delta x = u_xt+ \frac{1}{2}a_xt^2\\ \\u_x=initial\ velocity\ of \ car\ in\ x\ direction = 0,a_x=acceleration=0\\\\\Delta x = u_xt\\\\u_x=\frac{\Delta x}{t}=\frac{69}{2} =34.5\ m/s](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%20u_xt%2B%20%5Cfrac%7B1%7D%7B2%7Da_xt%5E2%5C%5C%20%5C%5Cu_x%3Dinitial%5C%20velocity%5C%20of%20%5C%20car%5C%20in%5C%20x%5C%20direction%20%3D%200%2Ca_x%3Dacceleration%3D0%5C%5C%5C%5C%5CDelta%20x%20%3D%20%20u_xt%5C%5C%5C%5Cu_x%3D%5Cfrac%7B%5CDelta%20x%7D%7Bt%7D%3D%5Cfrac%7B69%7D%7B2%7D%20%3D34.5%5C%20m%2Fs)
b)The car is moving at a constant speed in the horizontal direction, hence the initial velocity = final velocity
![v_x=u_x=34.5\ m/s\\\\v_y=u_y+a_yt\\\\v_y=0+gt\\\\v_y=10(2)=20\ m/s\\\\v=\sqrt{v_x^2+v_y^2}=\sqrt{34.5^2+20^2}=39.9\ m/s\\ v=39.9\ m/s](https://tex.z-dn.net/?f=v_x%3Du_x%3D34.5%5C%20m%2Fs%5C%5C%5C%5Cv_y%3Du_y%2Ba_yt%5C%5C%5C%5Cv_y%3D0%2Bgt%5C%5C%5C%5Cv_y%3D10%282%29%3D20%5C%20m%2Fs%5C%5C%5C%5Cv%3D%5Csqrt%7Bv_x%5E2%2Bv_y%5E2%7D%3D%5Csqrt%7B34.5%5E2%2B20%5E2%7D%3D39.9%5C%20m%2Fs%5C%5C%20v%3D39.9%5C%20m%2Fs)
I believe the correct gravity on the moon is 1/6 of Earth.
Take note there is a difference between 1 6 and 1/6.
HOWEVER, we should realize that the trick here is that the
question asks about the MASS of the astronaut and not his weight. Mass is an
inherent property of an object, it is unaffected by external factors such as
gravity. What will change as the astronaut moves from Earth to the moon is his
weight, which has the formula: weight = mass times gravity.
<span>Therefore if he has a mass of 50 kg on Earth, then he will
also have a mass of 50 kg on moon.</span>
Answer:
![c=71.4m/s](https://tex.z-dn.net/?f=c%3D71.4m%2Fs)
![\theta=8.54\textdegree](https://tex.z-dn.net/?f=%5Ctheta%3D8.54%5Ctextdegree)
Explanation:
From the question we are told that
Initial velocity of 60 m/s
Wind speed ![V_w= 15 m/s \angle 45 \textdegree](https://tex.z-dn.net/?f=V_w%3D%2015%20m%2Fs%20%5Cangle%20%2045%20%5Ctextdegree)
Generally Resolving vector mathematically
![sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6](https://tex.z-dn.net/?f=sin%2845%5Ctextdegree%2915%3D10.6%5C%5Ccos%2845%5Ctextdegree%2915%3D10.6)
Generally the equation Pythagoras theorem is given mathematically by
![c^2=a^2+b^2](https://tex.z-dn.net/?f=c%5E2%3Da%5E2%2Bb%5E2)
![c^2=10.6^2 +(10.6+60)^2](https://tex.z-dn.net/?f=c%5E2%3D10.6%5E2%20%2B%2810.6%2B60%29%5E2)
![c=\sqrt{10.6^2 +(10.6+60)^2}](https://tex.z-dn.net/?f=c%3D%5Csqrt%7B10.6%5E2%20%2B%2810.6%2B60%29%5E2%7D)
Therefore Resultant velocity (m/s)
![c=71.4m/s](https://tex.z-dn.net/?f=c%3D71.4m%2Fs)
b)Resultant direction
Generally the equation for solving Resultant direction
![\theta=tan^-1(\frac{y}{x})](https://tex.z-dn.net/?f=%5Ctheta%3Dtan%5E-1%28%5Cfrac%7By%7D%7Bx%7D%29)
Therefore
![\theta=tan^-1(\frac{10.6}{70.6})](https://tex.z-dn.net/?f=%5Ctheta%3Dtan%5E-1%28%5Cfrac%7B10.6%7D%7B70.6%7D%29)
![\theta=8.54\textdegree](https://tex.z-dn.net/?f=%5Ctheta%3D8.54%5Ctextdegree)
Answer:
500 N
Explanation:
Since the work done on the spring W = Fx where F = force applied and x = compression length = 0.170 m (since the spring will be compressed its full length when the force is applied)
Since W = 85.0 J and we need to find F,
F = W/x
= 85.0 J/0.170 m
= 500 N
So, the magnitude of force must you apply to hold the platform stationary at the final distance given above is 500 N.
Atomic number is equal to the number of protons and electrons
Atomic mass - protons = neutrons
protons + neutrons = atomic mass
I hope this helps