The weight of a 2959.0-kg car that is moving at a speed of 21 m / s would be 28998.2 Newtons, here speed of the car is the redundant value that is not used to calculate the weight of the car.
It can be defined as the force by which a body attracts another body toward its center as the result of the gravitational pull of one body and another.
As given in the problem, we have to determine the weight of a 2959.0-kg car that is moving at a speed of 21 m / s .
The weight of the car = mass of the car × acceleration due to gravity
= 2959 × 9.8
= 28998.2 Newtons
Thus, the weight of a 2959.0-kg car that is moving at a speed of 21 m / s would be 28998.2 Newtons.
To learn more about gravity here, refer to the link given below ;
#SPJ2
weight=mass*gravity
w=2959.0*10
w=29590N
6 seconds, Frequency=
Answer:
4 times/second
Explanation:
24/6 = 4
THE END
In this physics problem, by using the equation of motion, it was found that the time taken for a free falling object (under gravity and ignoring air resistance) to fall the second half of its total distance (t2) is sqrt(3) times greater than the time taken to fall the first half (t1).
This question falls under the category of Physics, and represents a common problem in the study of kinematics. To answer, we first define the two halves of the journey of the falling rock. We're given that the rock covers the first half distance in t1 time and the second half distance in t2 time. Since the falling object is under the influence of gravity (ignoring air resistance), it accelerates during its descent. Therefore, t2 > t1 or t2/t1 > 1.
A classical physics equation we can use is s = ut + 0.5at^2, where 's' represents distance, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time. Since the rock is dropped, the initial velocity 'u' is 0. Hence, the equation become 's = 0.5at^2'
Since the rock is being dropped, it falls under constant acceleration due to gravity (approximated as 9.8m/s^2). For the first half of the drop, the distance is s/2 = 0.5*g*(t1)^2, and for the second half of the drop, the distance is s/2 = 0.5*g*(t2)^2.
Solving these equations can help us find the relationship between t1 and t2. If we solve for t2 in terms of t1, keeping in mind that total distance (s) and acceleration due to gravity (g) remain the same in both cases, we obtain t2 = sqrt(3)*t1 approximately, i.e., t2/t1 = sqrt(3).
#SPJ12
Answer:
-you have to make yourse;f as light as possible so toss your bag, jacket, and shoes.
-Try to take a few steps backwards.
-Keep your arms up and out of the quicksand.
-Try to reach for a branch or person's hand to pull yourself out.
-Take deep breaths.
-Move slowly and deliberately.