Answer:
V=4.7m/s
Explanations:
Let Ma mass of cat A=7kg
Va velocity of cat A=7m/s
Mb mass of cat b=6.1kg
VB velocity of cat b=2m/s
From conservation of linear momentum
MaVa+MbVb=(Ma+Mb)V
7*7+6.1*2=(7+6.1)V
61.2=13.1V
V=4.7m/s
Answer:
The load has a mass of 2636.8 kg
Explanation:
Step 1 : Data given
Mass of the truck = 7100 kg
Angle = 15°
velocity = 15m/s
Acceleration = 1.5 m/s²
Mass of truck = m1 kg
Mass of load = m2 kg
Thrust from engine = T
Step 2:
⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:
T = (m1+m2)*g*sinθ
⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .
Resultant force on truck is F = T – m1*gsinθ
F causes the acceleration of the truck: F= m*a
This gives the equation:
T – m1*gsinθ = m1*a
T = m1(a + gsinθ)
Combining both equations gives:
(m1+m2)*g*sinθ = m1*(a + gsinθ)
m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ
m2*g*sinθ = m1*a
Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:
m2*g*sinθ = (7100 – m2)*a
m2*g*sinθ = 7100a – m2a
m2*gsinθ + m2*a = 7100a
m2* (gsinθ + a) = 7100a
m2 = 7100a/(gsinθ + a)
m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)
m2 = 2636.8 kg
The load has a mass of 2636.8 kg
Answer:
450N
Explanation:
Given data
Mass m= 75kg
Acceleration= 6m/s^2
From the Newtons first law, F=ma
substitute
F=75*6
F= 450N
Hence the force is 450N
Answer:
Fuel oil is mixed with a proper portion of compressed air
Explanation:
A gas turbine has three main part, which are
- combustion chamber
- air compressor
- power turbine
The combustion chamber is responsible for mixing fuel with a proper portion of compressed air.
The air compressor supplies air in sufficient quantity to satisfy the requirements of the combustion chamber
The power turbine produces the power to drive the air compressor.
Answer:
10.6cm
Explanation:
We are given 5.3cm below the starting point (spring extension).
Therefore, to find static vertical equilibrium, we use the equation:
kx = mg
Where:
k = spring constant =
=mg/5.3 kg/s²
We are told the object was dropped from rest.
Therefore:
loss in potential energy = gain in spring p.e
Let's use the expression:
mgx = ½kx²
We are asked to find the stretch at maximum elongation x.
To find x, we make x subject of the formula.
Therefore, we have:
x = 2mg/k (after rearranging the equation above)
x = (2mg) / (mg/5.3)
x = 10.6cm
