What is the minimum frequency of light necessary to emit electrons from titanium via the photoelectric effect?
2 answers:
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Answer:
Option B, 93 cm
Explanation:
An diagram of the seed's motion is attached to this solution.
This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.
And this would be obtained from the equations of motion,
First of, the height of the plant is related to some quantities of the motion with this relation.
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)
(substituting the t = √(2H/g) derived from above
R = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2
R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.
QED!
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.
Answer:
Explanation:
Given that:
mass of object A,
mass of object B,
speed of object A,
So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.
Answer:
vₓ = 20 m/s, v_{y} = -15 m / s
Explanation:
This is a conservation of moment problem, since it is a vector quantity we can work each axis independently
The system is formed by the two drones, so the forces during the crash are internal and the moment is conserved
X axis
Initial moment. Before the crash
p₀ = m₁ v₀ₓ + m₂ v₀ₓ
Final moment. After the crash
p_{fx} = (m₁ + m₂) vₓ
p₀ₓ =
m₁ v₀ₓ + m₂ v₀ₓ = (m₁ + m₂) vₓ
vₓ = (m₁ + m₂) v₀ₓ / (m₁ + m₂)
vₓ = v₀ₓ = 20 m/s
Y Axis
Initial
p_{oy} = m₁ v_{oy}
Final
p_{fy} = (m₁ + m₂) v_{y}
p_{oy} = p_{fy}
the drom rises and when it falls it has the same speed because there is no friction v_{oy} = -60 m/s
m₁ = (m₁ + m₂) v_{y}
v_{y} = m₁ / (m₁ + m₂) v_{oy}
v_{y} = 1/4 60
v_{y} = -15 m / s
Vertical speed is down