I think its b im not sure im in 8th grade
Answer:
D
Explanation:
I hope you get a good grade!
Answer:
work done lifting the bucket (sand and rope) to the top of the building,
W=67.46 Nm
Explanation:
in this question we have given
mass of bucket=20kg
mass of rope=![.2\frac{kg}{m}](https://tex.z-dn.net/?f=.2%5Cfrac%7Bkg%7D%7Bm%7D)
height of building= 15 meter
We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand
work done in lifting the rope is given as,
=
..............(1)
=
=22.5 Nm
work done in lifting the sand is given as,
.................(2)
Here,
F=mx+c
here,
c=20-18
c=2
m=![\frac{20-18}{15-0}](https://tex.z-dn.net/?f=%5Cfrac%7B20-18%7D%7B15-0%7D)
m=.133
Therefore,
![F=.133x+2](https://tex.z-dn.net/?f=F%3D.133x%2B2)
Put value of F in equation 2
![W_{2}=\int\limits^{15}_0 (.133x+2) \, dx](https://tex.z-dn.net/?f=W_%7B2%7D%3D%5Cint%5Climits%5E%7B15%7D_0%20%28.133x%2B2%29%20%5C%2C%20dx)
![W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm](https://tex.z-dn.net/?f=W_%7B2%7D%3D.133%20%5Ctimes%20112.5%2B2%5Ctimes15%5C%5CW_%7B2%7D%3D14.96%2B30%5C%5CW_%7B2%7D%3D44.96%20Nm)
Therefore,
work done lifting the bucket (sand and rope) to the top of the building,![W=W_{1}+W_{2}](https://tex.z-dn.net/?f=W%3DW_%7B1%7D%2BW_%7B2%7D)
W=22.5 Nm+44.96 Nm
W=67.46 Nm
Answer:
I'm pretty sure the answer is 0 m/s²
Explanation:
The horizontal velocity of the second rock is 5 m/s, so if we pretend air resistance doesn't exist, it will maintain that horizontal velocity, meaning that there is no horizontal acceleration.
Answer:
41°
Explanation:
Kinetic energy at bottom = potential energy at top
½ mv² = mgh
½ v² = gh
h = v²/(2g)
h = (2.4 m/s)² / (2 × 9.8 m/s²)
h = 0.294 m
The pendulum rises to a height of above the bottom. To determine the angle, we need to use trigonometry (see attached diagram).
L − h = L cos θ
cos θ = (L − h) / L
cos θ = (1.2 − 0.294) / 1.2
θ = 41.0°
Rounded to two significant figures, the pendulum makes a maximum angle of 41° with the vertical.