The force exerted on the car during the stop measures 6975 N.
Explanation: Given that the mass (m) is 930 kg, speed (s) at 56 km/h converts to 15 m/s, and the stopping time (t) is 2 s, we compute the force using F = m * a. Here, acceleration (a) can be obtained through a = s/t. The total force calculation confirms that F = 930 kg * (15 m/s) / 2 s results in 6975 N.
Answer:
The answer to your inquiry is Mass = 41230.7 g or 41.23 kg.
Explanation:
Data
Density = 0.737 g/ml
Mass = ?
Volume = 14.9 gal
1 gal = 3.78 l
Process
1.- Convert gallons to liters
1 gal ---------------- 3.78 l
14.9 gal ------------- x
x = 56.44 l
2.- Convert liters to milliliters
1 l ------------------- 1000 ml
56.44 l --------------- x
x = (56.44 x 1000) / 1
x = 56444 ml
3.- Calculate the mass
Formula
Density = 
Solving for mass
Mass = density x volume
Substituting values
Mass = 0.737 x 56444
Result
Mass = 41230.7 g or 41.23 kg.
Answer:
Height (h) = 17 m
Velocity (v) = 18.6 m/s
Explanation: This problem can be solved using kinematic motion equations.
Given Data
Initial velocity (u) = 0
Acceleration (a) = g
Time (t) = 1.9 seconds
First, we calculate the height.

Then, we find the final velocity

The acceleration graph is a linear representation described by y=9.8, as it remains constant:
The velocity graph can be represented by y=9.8x (where y signifies velocity and x indicates time):
The displacement graph can be described as y=4.9x^2 (with x as time and y as displacement):
These graphs apply exclusively from x=0 to x=1.9, so disregard other sections of the graphs.
The electric force between two objects is expressed as being proportional to the product of their charges and inversely proportional to the square of the distance separating them. In this instance, the distance between the first two charges is 19 cm. We formulate the equation k q1 q3/ (x)^2 = k q2 q3/ (19-x)^2, where x denotes the separation between q1 and q3. The charge q3 cancels out, and q2 is used in absolute terms. The resulting value of x is 5.79 cm.
Answer:
The initially bent young tree has been straightened by adjusting the tensions of the three guy wires to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Please calculate the force and moment reactions at the trunk's base point O, disregarding the weight of the tree.
C and D are situated 3.1' from the y-axis, while B and C are located 5.4' from the x-axis, and A has a height of 5.2'.
Explanation:
Refer to the attached image.