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:
28.3 m/s
Explanation:
From the question given above, the following data were obtained:
Angle of projection (θ) = 30°
Maximum height (H) = 10 m
Acceleration due to gravity (g) = 10 m/s²
Initial velocity (u) =?
Thus, we can obtain the minimum velocity cannon ball by using the following formula:
H = u²Sine² θ / 2g
10 = u² × (Sine 30)² / 2× 10
10 = u² × (0.5)² / 20
10 = u² × 0.25 / 20
10 = u² × 0.0125
Divide both side by 0.0125
u² = 10/ 0.0125
u² = 800
Take the square root of both side
u = √800
u = 28.3 m/s
Therefore, the minimum speed of the cannon ball is 28.3 m/s
The deeper you go, the more rock must be supported so the more force is required and the pressure goes up.
Answer:
2683.3N
Explanation:
According to coulombs law which states that "the force of attraction existing between two charge q1 and q2 is directly proportional to the product of the charges and inversely proportional to the square of the distance (d) between them. Mathematically |F|= k|q1| |q2| /d² where;
F is the force of attraction between the charges
q1 and q2 are the charges
d is the distance between them
k is the coulombs constant
Given |q1|= 38.9 × 10^-6C and |q2| = 27.6399×10^-6C d = 6cm = 0.06m
k = 8.98755 × 109 Nm² /C²
Substituting the given data's in the equation we have;
|F| = 8.98755 × 10^9×38.9×10^-6×27.6399×10^-6/0.06²
|F| = 9.66/0.06²
|F| = 9.66/0.0036
|F| = 2683.3N
The magnitude of the force will be 2683.3N
Note that the modulus of the charges changes negative value of q2 to positive value. The opposite signs of the charges doesn't affect the final calculation, it only tells the force of attraction or repulsion between the charges. Since they are unlike charges, they will attract each other in the field.
It is B
When we look at a red apple, it's absorbing colors from the sunlight. It absorbs all the colors of the rainbow—except for red. The red light reflects off the apple and our brain and eyes work together to let us know what color we are seeing.
