Answer:
It should be mentioned that an efficient way to work this vector addition problem is with the cosine law for general triangles (and since
a
,
b
and
r
from an isosecles triangle, the angles are easy to figure). However, in the interest of reinforcing the usual systematic approach to vector addition, we note that the angle
b
makes with the +x axis is 30
0
+105
0
=135
0
(a) The x component of
r
x
=(10.0m)cos30
0
+(10.0m)cos135
0
=1.59m
(b) the y component of
r
is r
y
=(10.0m)sin135
0
= 12.1 m.
(c) the magnitude of
r
is r = ∣
r
∣=
(1.59m)
2
+(12.1m)
2
=12.2m
(d) The angle between
r
and the +x direction is tan
−1
[(12.1m)/(1.59m)]=82.5
0
Explanation:
Assuming you are supposed to write each conversion in scientific notation:
(2) 1 m = 100 cm, so
(67 cm) × (1/100 m/cm) = 67/100 m = 0.67 m = 6.7 × 10 ⁻¹ m
(3) 1 km = 1,000 m, so
(1.2 km) × (1000 m/km) = 1200 m = 1.2 × 10³ m
(4) 1 m = 1,000 mm = 10³ mm, so
(6.2 × 10 ⁻³ m) × (10³ mm/m) = 6.2 mm
(5) 1 m = 1,000,000,000 nm = 10⁹ nm, so
(4.05 × 10³ nm) × (1/10⁹ m/nm) = 4.05 × 10 ⁻⁶ m
(6) 1 g = 1,000,000 µg = 10⁶ µg, so
(3200 µg) × (1/10⁶ g/µg) = 3200 × 10 ⁻⁶ g = 3.2 × 10 ⁻³ g
Answer:
2.06 m at 49.7 or 50 degree South of East
Explanation:
take x axis and y axis and follow as explained in attachment
The decrease in gravitational potential energy of the system is given by
![\Delta U = (mg) \Delta h](https://tex.z-dn.net/?f=%20%5CDelta%20U%20%3D%20%28mg%29%20%5CDelta%20h%20)
where
m is the mass, g is the gravitational acceleration, and
is the variation of height of the system.
also corresponds to the weight of the diver, therefore if we rearrange the equation and we use
and
, we can find her weight:
![mg=\frac{\Delta U}{\Delta h}=\frac{28000 J}{32 m}=875 N](https://tex.z-dn.net/?f=%20mg%3D%5Cfrac%7B%5CDelta%20U%7D%7B%5CDelta%20h%7D%3D%5Cfrac%7B28000%20J%7D%7B32%20m%7D%3D875%20N%20%20%20)
A body accelerates due to net force acting on it. As newton's second law of motion states:
when a net force acts on a body, it produces acceleration in the body in the direction of the net force