Answer:
The scale will read less than 700 N
Explanation:
Given;
normal weight of the person, W = 700 N
The upward acceleration of the elevator is given by Newton's second law of motion;
F = ma
Also, the weight of the person, W = mg
Net force on the person when the elevator accelerates upward is given as;
R = ma + mg
When the elevator slows down at a steady rate, then net force on the person is given as;
R = mg - ma
R = m(g - a), this net force on the person is also the reading of the scale.
Thus, the scale will read less than 700 N.
Answer:
e = 0.46 m
Explanation:
From the laws of friction, frictional force, F is proportional to normal reaction, R.
F₁ = μR
where μ is coefficient of friction; R = mg and g = 9.8 ms⁻²
Also, from Hooke's law, extension, e, in an elastic spring is proportional to applied force.
F₂ = Ke
where K is force constant of the spring
Since the box is just about to move, the coefficient of friction involved is static friction.
The force on the spring equals the frictional force experienced by the box the box; F₁ = F₂
Ke = μR
e = μR/K
where μ = 0.65; R = 18 kg * 9.8 ms⁻²; K = 250 N/m
e = (0.65 * 18 * 9.8)/250
e = 0.46 m
The correct answer is :
B. the value of q is positive.
Hope this helps,
Davinia.
Answer:
m=0.5kg
h = 180 cm =1.8 mh=180cm=1.8m
Initial potential energy of the object is:
E_p=m*g*hE
p
=m∗g∗h
Kinetic energy at the surface:
E_k=\frac{mv^2}{2}E
k
=
2
mv
2
According to the law of conservation of energy (assuming no air resistance):
E_p = E_kE
p
=E
k
mgh=\frac{mv^2}{2}mgh=
2
mv
2
Solving for v:
v=\sqrt{2gh}v=
2gh
p=mvp=mv
So,
p= m*v = m\sqrt{2gh}p=m∗v=m
2gh
Calculating:
p= 0.5\sqrt{2*9.8*1.8}\approx 2.97 \frac{kg*m}{s}p=0.5
2∗9.8∗1.8
≈2.97
s
kg∗m
Answer:
p \approx 2.97 \frac{kg*m}{s}p≈2.97
s
kg∗m
The general trend is that electronegativity increases as you go up and to the right on the periodic table