Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
Answer:
350.72 m/s
Explanation:
Formula for velocity of wave is;
v = fλ
Where;
v is speed
f is frequency
λ is wavelength
We are given;
f = 512 Hz
λ = 0.685 m
Thus;
v = 512 × 0.685
v = 350.72 m/s
Answer:
Explanation: Covalent bonding occurs when pairs of electrons are shared by atoms. Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outermost (valence) electrons, atoms can fill up their outer electron shell and gain stability.
Answer:
The amount of current that must flow through the wire for it to be suspended against gravity by magnetic force = 6.125 A
Explanation:
Force on a wire carrying current in an electric field is given by
F = (B)(I)(L) sin θ
For this question,
The magnetic force must match the weight of the wire.
F = mg
mg = (B)(I)(L) sin θ
(m/L)g = (B)(I) sin θ
Mass per unit length = 75 g/m = 0.075 kg/m
B = magnetic field = 0.12 T
I = ?
g = acceleration due to gravity = 9.8 m/s
θ = angle between wire's current direction and magnetic field = 90°
0.075 × 9.8 = 0.12 × I sin 90°
I = 0.075 × 9.8/0.12 = 6.125 A
Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.
So we have 1/2 MV^2 = MGH
V^2 = 2GH
V = âš2GH
V = âš( 2 * 9.8 * 325)
V = âš 6370
V = 79.81 m/s
