Answer:
V = 48.4m/s, θ = 73.6° below the horizontal.
Explanation:
Given h = 109.53m, x = range = 65m,
θ = 0°
This problem involves the concepts of projectile motion.
Let yo = h = 109.53m
y = final position on the y axis. = 0m (ground level)
y = yo + vosinθ – 1/2gt²
0 = 109.53 + vosin0° – 1/2×9.8×t²
0 = 109.53 – 4.9t²
4.9t² = 109.53
t² = 109.53/4.9 = 22.35
t = √22.35 = 4.73s
So it takes the ball t = 4.73 seconds to get to the ground from the launch point.
x = (vocosθ)×t
65 = (vocos0°)×4.73
65/4.73 = vo
Vo = 13.7m/s
Vx = Vox = Vocosθ = 13.7cos0° = 13.7m/s
Vy = Voy – gt = Vosinθ – gt
Vy = 13.7sin0° – 9.8×4.73 = –46.4m/s
V = √(Vy² + Vx²) = √(-46.4² + 13.7²)
V = 48.4m/s
θ = Tan-¹(vy/vx) = tan-¹(-46.4/13.7) = -73.6°
θ = 73.6° below the horizontal
V = 48.4m/s
Answer:
609547.12 Pa ≈ 6.10×10^5 Pa
Explanation:
Step 1:
Data obtained from the question. This include the following:
Force (F) = 49.8 N
Radius (r) = 0.00510 m
Pressure (P) =..?
Step 2:
Determination of the area of the head of the nail.
The head of a nail is circular in nature. Therefore, the area is given by:
Area (A) = πr²
With the above formula we can obtain the area as follow:
Radius (r) = 0.00510 m
Area (A) =?
A = πr²
A = π x (0.00510)²
A = 8.17×10^-5 m²
Therefore the area of the head of the nail is 8.17×10^-5 m²
Step 3:
Determination of the pressure exerted by the hammer.
This is illustrated below:
Force (F) = 49.8 N
Area (A) = 8.17×10^-5 m²
Pressure (P) =..?
Pressure (P) = Force (F) /Area (A)
P = F/A
P = 49.8/8.17×10^-5
P = 609547.12 N/m²
Now, we shall convert 609547.12 N/m² to Pa.
1 N/m² = 1 Pa
Therefore, 609547.12 N/m² = 609547.12 Pa.
Therefore, the pressure exerted by the hammer on the nail is 609547.12 Pa or 6.10×10^5 Pa
Answer:
0.3 seconds approximately
Explanation:
Given that the height of the building is 55.9 m and cement block accidentally falls from rest from the edge of a 55.9-m-high building. The initial velocity of the block will be equal to zero. The final velocity will be achieved by using the formula:
V^2 = U^2 + 2gH
Where g = 9.8 m/s^2
H = 55.9m
V^2 = 0 + 2 × 9.8 × 55.9
V^2 = 1095.64
V = sqrt(1095.64)
V = 33.1 m/s
The velocity When the block is 11.2 m above the ground will be 33.1 m/s and the height h = 11.2 - 1.7 = 9.5 m
The time of escape can be calculated by using the formula;
h = Ut + 1/2gt^2
9.5 = 33.1t + 1/2 × 9.8 × t^2
9.5 = 33.1t + 4.9t^2
4.9t^2 + 33.1t - 9.5
By using quadratic formula to calculate the time, the negative value is ignored and the positive values will be the time the man has to get out of the way.
Please find the attached file for the remaining solution.
Answer:
<h2>1.5 ohms</h2>
Explanation:
Power is expressed as P = V²/R
R = resistance
V = supplied voltage
Given P = 600W and V = 30V
R = V²/P
R = 30²/600
R = 900/600
R = 1.5ohms
magnitude of its resistance is 1.5ohms
Answer:
The mass of the body is 2.2 kg.
Explanation:
Given that,
Force acting on a body 1,
Force acting on a body 2,
Direction,
Acceleration,
To find,
Mass of the body.
Solution,
Let F is the net force acting on the body. It is given by :
F = 43.22 N
Second law of motion is given by :
F = ma
m = 2.16 kg
or
m = 2.2 kg
So, the mass of the body is 2.2 kg.