First make sure you draw a force diagram. You should have Fn going up, Fg going down, Ff going left and another Fn going diagonally down to the right. The angle of the diagonal Fn (we'll call it Fn2) is 35° and Fn2 itself is 80N. Fn2 can be divided into two forces: Fn2x which is horizontal, and Fn2y which is vertical. Right now we only care about Fn2y.
To solve for Fn2y we use what we're given and some trig. Drawing out the actual force of Fn2 along with Fn2x and Fn2y we can see it makes a right triangle, with 80 as the hypotenuse. We want to solve for Fn2y which is the opposite side, so Sin(35)=y/80. Fn2y= 80sin35 = 45.89N
Next we solve for Fg. To do this we use Fg= 9.8 * m. Mass = 30kg, so Fg = 9.8 * 30 = 294N.
Since the chair isn't moving up or down, we can set our equation equal to zero. The net force equation in the vertical direction will be Fn + Fn2y -Fg = 0. If we plug in what we know, we get Fn + 45.89 -294 = 0. Then solve this algebraically.
Fn +45.89 -294 = 0
Fn +45.89 = 294
Fn = 248.11 N
You'll get a more accurate answer if you don't round Fn2y when solving for it, it would be something along the lines of 45.88611 etc
<span>the formula q = 375 g * 25 C * 4.186 J / (g*C) = 39,243.75 J q represents the heat in Joules , m the mass in grams, difference of temperature in Celsius degree, and 4.186 J/(g*C) is the specific heat of water( I assume the water is in liquid from and will remain liquid). Approximately 39.24 kJ once you round and transform to kJ..1 kJ=1000J</span>
Answer:
B. no
Explanation:
- When any body moves at a speed comparable to the speed of light (<em>i.e. relativistic speed</em>) then the observer sees a contraction in length of the body along the axis of motion.
Assuming that the motion of the body is along the axis of the rod, then the observer will measure its length to be lesser than its length at rest.
<u>Then according to Einstein's theory of relativity:</u>
where:
original length of the object (along the direction of motion)
L = observed length of the rod
Lorentz factor
Neglecting air resistance, the horizontal component remains constant. The angle doesn't matter.
Answer:
The magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J
Explanation:
Given;
radius of the circular loop of wire = 0.5 m
current in circular loop of wire = 2 A
strength of magnetic field in the wire = 0.3 T
τ = μ x Bsinθ
where;
τ is the magnitude of the magnetic torque
μ is the dipole moment of the magnetic field
θ is the inclination angle, for a plane area perpendicular to the magnetic field, θ = 90
μ = IA
where;
I is current in circular loop of wire
A is area of the circular loop = πr² = π(0.5)² = 0.7855 m²
μ = 2 x 0.7885 = 1.571 A.m²
τ = μ x Bsinθ = 1.571 x 0.3 sin(90)
τ = 0.4713 J
Therefore, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J
