We don't see any circuit diagrams.
This worries us for a few seconds, until we realize that we don't know anything about the experiment described in the problem either, so we don't have to worry about it at all.
Answer:
fcosθ + Fbcosθ =Wtanθ
Explanation:
Consider the diagram shown in attachment
fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)
Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)
Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)
sum of x-direction forces = 0
fx+ Fbx=Wx
fcosθ + Fbcosθ =Wtanθ
#1
Volume of lead = 100 cm^3
density of lead = 11.34 g/cm^3
mass of the lead piece = density * volume
so its weight in air will be given as
now the buoyant force on the lead is given by
now as we know that
so by solving it we got
V = 11.22 cm^3
(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N
(iii) Buoyant force = 0.11 N
(iv)since the density of lead block is more than density of water so it will sink inside the water
#2
buoyant force on the lead block is balancing the weight of it
(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N
(iii) Buoyant force = 11.11 N
(iv) since the density of lead is less than the density of mercury so it will float inside mercury
#3
Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid
The number of significant digits of any measurement is determined by the instrument used for such measurement. For example, in this case, we have the height of a small child being measured. We can use a simple ruler for this, and we see that a ruler has ten divisions for 1 cm. This means that the ruler cannot measure beyond the size of 0.1 cm or 1 mm. Hence, when we report the height of the small child, we report it to one significant digit after the decimal place. As an example, if we measure a child's height to be 90 full cm divisions and 8 smaller divisions, we report it as 90.8 cm but not 90.83 or 90.86 cm.
Answer:
a) a= 8.33 m/s², T = 12.495 N
, b) a = 2.45 m / s²
Explanation:
a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.
We apply Newton's second law to the lower charge
fr₁ + fr₂ = ma
The equation for the force of friction is
fr = μ N
Y Axis
N - W₁ –W₂ = 0
N = W₁ + W₂
N = (m₁ + m₂) g
Since the beams are the same, it has the same mass
N = 2 m g
We replace
μ₁ 2mg + μ₂ mg = m a
a = (2μ₁ + μ₂) g
a = (2 0.30 + 0.25) 9.8
a= 8.33 m/s²
Let's look for cable tension with beam 2
T = m₂ a
T = 1500 8.33
T = 12.495 N
b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal
fr = m₂ a₂
N- w₂ = 0
N = W₂ = mg
μ₂ mg = m a₂
a = μ₂ g
a = 0.25 9.8
a = 2.45 m / s²
