Answer:
a = 0.16
Explanation:
given,
mass of the object 1 = 0.2 kg
mass of the object 2 = 0.3 kg
acceleration when force is on 0.2 kg = 0.4 m/s²
acceleration when both mass are combine = ?
F = m a
F = 0.2 × 0.4
F = 0.08 N
force acting is same and total mass = 0.2 + 0.3 = 0.5 Kg
F = m a
a = 0.16 m/s²
the acceleration acting when both the body is attached is a = 0.16
There are other forces at work here nevertheless we will imagine
it is just a conservation of momentum exercise. Also the given mass of the
astronaut is light astronaut.
The solution for this problem is using the formula: m1V1=m2V2 but
we need to get V1:
V1= (m2/m1) V2
V1= (10/63) 12 = 1.9 m/s will be the final speed of the astronaut after
throwing the tank.
Answer:
v_f = 17.4 m / s
Explanation:
For this exercise we can use conservation of energy
starting point. On the hill when running out of gas
Em₀ = K + U = ½ m v₀² + m g y₁
final point. Arriving at the gas station
Em_f = K + U = ½ m v_f ² + m g y₂
energy is conserved
Em₀ = Em_f
½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂
v_f ² = v₀² + 2g (y₁ -y₂)
we calculate
v_f ² = 20² + 2 9.8 (10 -15)
v_f = √302
v_f = 17.4 m / s
Answer:
There is 148.35 Joules of heat is released in the process.
Explanation:
Given that,
Heat capacity of the object, 
Initial temperature, 
Final temperature, 
We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :
Let the mass of the object is 10 g or 0.01 kg
So,
Q = 148.35 Joules
So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.
<span> Rising, warm, moist air masses cool and release precipitation as they rise and then at high altitude, cool
and sink back to the surface as dry air masses after moving north or south of the tropics.
</span>
