Answer:
They are both correct.
Explanation:
The density of an object is defined as the ratio of its mass to its volume. This implies that the density of the object is both proportional to the mass and also to the volume of the object. John only mentioned mass which is correct. Linda mentioned the second variable on which density depends which is the volume of the object.
Hence considering the both statements objectively, one can say that they are both correct.
Answer:
1977.696 J
Explanation:
Given;
Weight of the box = 28.0 kg
Force applied by the boy = 230 N
angle between the horizontal and the force = 35°
Therefore,
the horizontal component of the force = 230 × cosθ
= 230 × cos 35°
= 188.405 N
Coefficient of kinetic friction, μ = 0.24
Force by friction, f = μN
here,
N = Normal force = Mass × acceleration due to gravity
or
N = 28 × 9.81 = 274.68 N
therefore,
f = 0.24 × 274.68
or
f = 65.9232 N
Now,
work done by the boy, W₁ = 188.405 N × Displacement
= 188.405 N × 30
= 5652.15 J
and,
the
work done by the friction, W₂ = - 65.9232 N × Displacement
= - 65.9232 N × 30 m
= - 1977.696 J
[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]
Explanation:
Using Kinematics,
we have a = (v - u) / t.
Therefore a = (36m/s - 22m/s) / 5s = 2.8m/s².
The law of universal gravitation says that one object attracts every other object using a proportional force to the mass of the object.
Answer:
E
= -4556.18 N/m
Explanation:
Given data
u = 3.6×10^6 m/sec
angle = 34°
distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in
Question)
from the projectile motion the horizontal distance traveled by electron is
x = u×cosA×t
⇒t = x/(u×cos A)
We also know that force in an electric field is given as
F = qE
q= charge , E= strength of electric field
By newton 2nd law of motion
ma = qE
⇒a = qE/m
Also, y = u×sinA×t - 0.5×a×t^2
⇒y = u×sinA×t - 0.5×(qE/m)×t^2
if y = 0 then
⇒t = 2mu×sinA/(qE) = x/(u×cosA)
Also, E = 2mu^2×sinA×cosA/(x×q)
Now plugging the values we get
E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})
E
= -4556.18 N/m
