B would be your right answer. Because Jupiter’s gravity pull is much stronger than earths
Answer:
See answers below
Explanation:
vf = vi + gt
0 = (120sin55°) + (-9.8)t
t = 12.2 s
12.2 x 2 = 24.4 s
x = (vi)t + (1/2)at²
x = (120cos55°)(24.4) + 0
x = 64.79 m
a.
vf = vi + gt
0 = (120sin55°) + (-1.6)t
t = 74.98 s
74.98 x 2 = 149.96 s
x = (vi)t + (1/2)at²
x = (120cos55°)(149.96) + 0
x = 398.18 m
Answer:
6.57 m/s
Explanation:
First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement
F=kx; F=180(.3) = 54 N
Next from Newton's second law find the acceleration of the mass.
Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²
Now use the kinematic equation for velocity (or speed)
v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.
v₀=0, since the mass is at rest before we release it
a=72 m/s² (from above)
x₀=0 as the start position already compressed
x₂=0.3m (this puts the spring back to it's natural length)
v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²
v₂=
= 6.57 m/s
Answer:
Explanation:
Question 1
An arrow weighing 20g shortly after firing has a speed of 50m / s. Calculate the work done by the athlete. What is the potential energy of the elasticity of the tensed string?
mass m = 20g = 20/1000 = 0.02kg
speed v = 50m / s
P.E = K.E = ½mv²
P.E = ½ × 0.02 × 50²
P.E = 25 J
work done = P.E = 25J
Qestion 2
A 80 kg athlete stood on a trampoline with a coefficient of elasticity of k = 2 kN / m. As far as the edge of the trampoline lowers.
force of elasticity
F = -kx
x = F / k
in our case F will be the force of pressure or gravity
F = mg
g is gravitational acceleration, and according to Newton's second law, acceleration is force through mass - unit of force N, unit of mass kg. Acceleration either in m / s ^ 2 or N / kg
F = 80kg * 10N / kg = 800 N
x = 800N / -2000N = -0.4
The trampoline will lower, so from the level by 0.4 meters and hence this minus
In adding measurements, we should remember that the units of the values to be added should be homogeneous which means all addends should have the same units. For this case, we must convert all the masses to units grams. The total mass of the of the all the masses weighed is:
1.21 kg ( 1000 g / 1 kg ) + 546 mg ( 1 g / 1000 mg ) + 23.14 g = 1233.686 grams
To correctly report this, it should be written in scientific notation which would be 1.23 x 10^3 g. In adding, the least number of significant figures should be followed. In this case, it is 3 significant figures.