Answer: Weight
Explanation:The mass of an object is a measure of the quantity of matter in the object. This quantity remains constant under any circumstance.
However, the same cannot be said about the weight of such object.
The weight of the object is very much dependent on the acceleration due to gravity which is a accelerational pull (by convention-- a pull downwards).
This is why an object tends to fall when it is thrown upwards on the earth for instance.
The statements above consequently infer that since the gravitational field of Jupiter is greater than that of the earth, the acceleration due to gravity on Jupiter is greater than that on earth.
And since the weight of an object(W) is a product of its mass and the acceleration due to gravity at that point.
Consequently, the object's weight on Jupiter would be greater than its weight on earth.
Please note; The Mass of the object remains constant everywhere.
Answer:
a) -113 N/C
b) -423.71 V
Explanation:
The Step by step explanation and solution has been attached in the attachment section.
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:
V = 10.88 m/s
Explanation:
V_i =initial velocity = 0m/s
a= acceleration= gsinθ-
cosθ
putting values we get
a= 9.8sin25-0.2cos25= 2.4 m/s^2
v_f= final velocity and d= displacement along the inclined plane = 10.4 m
using the equation
v_f= 7.04 m/s
let the speed just before she lands be "V"
using conservation of energy
KE + PE at the edge of cliff = KE at bottom of cliff
(0.5) m V_f^2 + mgh = (0.5) m V^2
V^2 = V_f^2 + 2gh
V^2 = 7.04^2 + 2 x 9.8 x 3.5
V = 10.88 m/s
