Answer:
Given the experimental setup shown below, where should you put object C, ... applies a force of 150 N. What is the mechanical advantage of the hammer? ... Explain why tin snips, designed for cutting metal, have long handles and short blades. ... advantage for scissors, a first-class lever, is the length of the handle divided ...
Explanation:
Answer:
The rate of change of distance between the two ships is 18.63 km/h
Explanation:
Given;
distance between the two ships, d = 140 km
speed of ship A = 30 km/h
speed of ship B = 25 km/h
between noon (12 pm) to 4 pm = 4 hours
The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =
140 km - 120 km = 20 km
(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)
The displacement of ship B at 4pm = 25 km/h x 4h = 100 km
Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;
r² = a² + b²
r² = 20² + 100²
r = √10,400
r = 101.98 km
The rate of change of this distance is calculated as;
r² = a² + b²
r = 101.98 km, a = 20 km, b = 100 km
Answer:C
Explanation:
Partially submerged block along with vessel is accelerated upwards .
Initially the block weight is supported by buoyant force such that it is in equilibrium.
when the system start accelerating upwards then the effective gravity will be
g+a where a is the acceleration of the system.
so only net gravity is increased so block will not ascend or descend.
Mathematically
where 
density of liquid
V=volume of object inside the water
To find the tangent plane to the surface f(x,y,z)=0 at a point (X,Y,Z) we use the following method:
<span>Calculate grad f = (f_x, f_y, f_z). The normal vector to the surface at the point (X,Y,Z) is grad f(X,Y,Z). The equation of a plane with normal vector n which passes through the point p is (r-p).n=0, where r=(x,y,z) is the position vector. So the equation of the tangent plane to the surface through the point (X,Y,Z) is ((x,y,z)-(X,Y,Z)).grad f(X,Y,Z)=0. </span>
<span>Now in your case we have f(x,y,z)=y-x^2-z^2, so grad f=(-2x,1,-2z), and the equation of the tangent plane at the point (X,Y,Z) is </span>
<span>((x,y,z)-(X,Y,Z)).(-2X,1,-2Z)=0, </span>
<span>that is </span>
<span>-2X(x-X)+1(y-Y)-2Z(z-Z)=0, </span>
<span>i.e. </span>
<span>-2Xx+y-2Zz = -2X^2+Y-2Z^2. (1) </span>
<span>Now compare this equation with the plane </span>
<span>x + 2y + 3z = 1. (2) </span>
<span>The two planes a_1x+b_1y+c_1z=d_1, a_2x+b_2y+c_2z=d_2 are parallel when (a_1,b_1,c_1) is a multiple of (a_2,b_2,c_2). So the two planes (1),(2) are parallel when (-2X,1,-2Z) is a multiple of (1,2,3), and we have </span>
<span>(-2X,1,-2Z)=1/2(1,2,3) </span>
<span>for X=-1/4 and Z=-3/4. On the paraboloid the corresponding y coordinate is Y=X^2+Z^2=1^4+9^4=5/2. </span>
<span>So the tangent plane to the given paraboloid at the point (-1/4,5/2,-3/4) is parallel to the given plane.</span>
