1.5 microseconds means . . .
-- 1.5 millionths of a second
-- 1.5 x 10⁻⁶ second
-- 0.0000015 second
Walking at a speed of 2.1 m/s, in the first 2 s John would have walked
(2.1 m/s) (2 s) = 4.2 m
Take this point in time to be the starting point. Then John's distance from the starting line at time <em>t</em> after the first 2 s is
<em>J(t)</em> = 4.2 m + (2.1 m/s) <em>t</em>
while Ryan's position is
<em>R(t)</em> = 100 m - (1.8 m/s) <em>t</em>
where Ryan's velocity is negative because he is moving in the opposite direction.
(b) Solve for the time when they meet. This happens when <em>J(t)</em> = <em>R(t)</em> :
4.2 m + (2.1 m/s) <em>t</em> = 100 m - (1.8 m/s) <em>t</em>
(2.1 m/s) <em>t</em> + (1.8 m/s) <em>t</em> = 100 m - 4.2 m
(3.9 m/s) <em>t</em> = 95.8 m
<em>t</em> = (95.8 m) / (3.9 m/s) ≈ 24.6 s
(a) Evaluate either <em>J(t)</em> or <em>R(t)</em> at the time from part (b).
<em>J</em> (24.6 s) = 4.2 m + (2.1 m/s) (24.6 s) ≈ 55.8 m
The pressure exerted by a fluid solely relies on the depth or height of the fluid, its density, and the gravitational constant. These three are related in the equation:
Pressure = density x g x height
In the problem, point A is within the block inside the tank. The water above the block is assumed to be 0.6 meters. This gives a point A pressure of:
P = 1000 kg/m^3 * 9.81 m/s^2 * 0.6 m = 5,886 Pa or 5.88KPa
Answer:
14.56 N
Explanation:
According to Newton's second law of motion,
The force acting on a body is equal to the product of its mass and acceleration acting on it.
Weight of a body on a planet is equal to the force due to gravity of the planet. So, weight is the product of mass of body and acceleration due to gravity of the planet.
Here, acceleration due to gravity on planet A is m/s² and mass of an object on it is kg.
So, weight of an object, , on planet A is given as,