Answer a is one of keplers 3 laws
Answer:
x coordinate = -1.66 m
y coordinate is = -0.825m
Explanation:
Suppose z be the distance form the first charge and z + sqrt(1^2 +.5^2) be the distance from the second So z + sqrt(1+.25) = z + 1.12
We have k*2.0x10^-6/s^2 = k*6x10^-6/(s+1.12)^2
0.0356s^2 -0.019s-0.0897=0
s=1.876m
The angle of the line between the two charges is arctan(.5/1) = 26.6o
x coordinate = -1.876*cos(26.6) = -1.66m
y coordinate is -1.876*sin(26.6) = -0.825m
Answer:
Explanation:
mass of the fellow ( m ) = 66kg
acceleration of fellow a
v = u + at
4.5 = 0 + a x 2
a = 4.5 /2
= 2.25 m / s²
Net force acting on fellow in upward direction by the surface of elevator
R - mg where R is reaction force of the surface of the elevator
Applying Newton's law of motion
R - mg = ma
R = m (g +a )
= 66 x ( 9.8 + 2 )
= 778.8 N
This will be the scale reading .
Answer: You didn't put what a, b, c, or d is so how is anyone supposed to know what the answer is?
Answer:
The work done by the gravel to stop the truck is 520.44 kJ
Explanation:
<u>Step 1</u>: Data given
Mass of the truck = 3047.8 kg
The ramp has an angle of 9.5 °
Velocity of the truck = 20.68 m/s
distance = 26.6 meters
<u>Step 2:</u> Calculate initial kinetic energy
sin 9.5° = 0.165
h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m
Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ = initial kinetic energy
<u>Step 3: </u>Calculate potential energy
Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ
<u>Step 4:</u> What work is done by the truck on the gravel?
Frictional energy Ef = 651.7 kJ - 131.26 kJ = 520.44 kJ
