Answer:
sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j
Explanation:
We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis
So x component of the vector 
y component of the vector 
So vector will be 6.06i+3.5j
Now other vector of length of 7 units and makes an angle of 120° with positive x-axis
So x component of vector 
y component of the vector 
Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j
Answer:
5694000 min
Explanation:
Let's suppose the average American watches 4 hours of TV every day. First, we will calculate how many minutes they watch per day. We will use the conversion factor 1 h = 60 min.
(4 h/day) × (60 min/1 h) = 240 min/day
They watch 240 minutes of TV per day. Now, let's calculate how many minutes they watch per year. We will use the conversion factor 1 year = 365 day.
240 min/day × (365 day/year) = 87600 min/year
They watch 87600 min/year. Finally, let's calculate how many minutes they spend watching TV in 65 years.
87600 min/year × 65 year = 5694000 min
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
Answer:
Option ( B ) is correct .
Explanation:
To lift a heavy weight , inclined plane is used . Use of inclined plane , makes the task easier because instead of force mg , force mg sinθ is to be used which is less than mg . Here θ is inclination of inclined plane.
If h be the height by which weight is to be lifted
potential energy acquired by weight = mgh
work done by force mg sinθ = mgsinθ x d where d is displacement required .
mg sinθ x d = mgh ( work done by force = potential energy stored in luggage )
d = h / sinθ
d will be more than h
Hence inclined plane increases the distance to be covered by force applied though it decreases the force itself.
Hence option ( B ) is correct .
