Find the area of the squares.
4 ×4 = 16
Find the area of the wall.
20 × 10 =200
Divide 200 by 16.
200 ÷ 16 = 12.5
You need 12.5 squares.
Answer:
Therefore,
![r=\sqrt[3]{\frac{3V}{4\pi }}](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%20%7D%7D)
is the required r
Step-by-step explanation:
Given:
Volume of inside of the sphere is given as
where r is the radius of the sphere
To Find:
r =?
Solution:
We have
......Given
![3\times V=4\pi r^{3} \\\\\therefore r^{3}=\frac{3V}{4\pi } \\\\\therefore r=\sqrt[3]{\frac{3V}{4\pi }} \textrm{which is the expression for r}](https://tex.z-dn.net/?f=3%5Ctimes%20V%3D4%5Cpi%20r%5E%7B3%7D%20%5C%5C%5C%5C%5Ctherefore%20r%5E%7B3%7D%3D%5Cfrac%7B3V%7D%7B4%5Cpi%20%7D%20%5C%5C%5C%5C%5Ctherefore%20r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%20%7D%7D%20%5Ctextrm%7Bwhich%20is%20the%20expression%20for%20r%7D)
Therefore,
is the required r
Work done on cart
:
W
=
6000
J
Kinetic energy of cart
:
E
K
=
6000
J
Explanation:
Work done
W
is defined as the product of force
F
and distance
s
:
⇒
W
=
F
⋅
s
Let's substitute the values of
F
and
s
into the equation:
⇒
W
=
300
N
⋅
20
m
⇒
W
=
300
kg
⋅
m s
−
2
⋅
20
m
⇒
W
=
300
⋅
20
kg
⋅
m
2
s
−
2
⇒
W
=
6000
kg
⋅
m
2
s
−
2
∴
W
=
6000
J
The gain in kinetic energy
E
K
is the same amount as the work done moving the cart:
∴
E
K
=
6000
first you do 8×12=92 then you do 92+2=94 94×12=1,128 so you save1,128 if I'm not wrong
Answer:
2.
i) (c)
ii) (d)
iii) (b)
iv) (a)
3.
ii) 80°
Step-by-step explanation:
2.
i) 2x+x+3x=180°
x=30°
ii) x+110=180
x=70
iii) x+2x+30°=180
x=50°
iv) 2x+15°+45°+x=180
3x+60°=180
x=40°
3.
ii) 2x+5+25=180°
2x+30°=180°
x=75°
the measure: x+5°= 75°+5°=80°
