Answer:
A body travels 10 meters during the first 5 seconds of its travel,and a total of 30 meters over the first 10 seconds of its travel
20miles / 5sec = 4miles /sec would be the average speed for the last 20 m
Explanation:
The answer is 4 m/s.
In the first 5 seconds, a body travelled 10 meters. In the first 10 seconds of the travel, the body travelled a total of 30 meters, which means that in the last 5 seconds, it travelled 20 meters (30m + 10m).
The relation of speed (v), distance (d), and time (t) can be expressed as:
v = d/t
We need to calculate the speed of the second 5 seconds of the travel:
d = 20 m (total 30 meters - first 10 meters)
t = 5 s (time from t = 5 seconds to t = 10 seconds)
Thus:
v = 20m / 5s = 4 m/s
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Answer:
The mass of the rule is 56.41 g
Explanation:
Given;
mass of the object suspended at zero mark, m₁ = 200 g
pivot of the uniform meter rule = 22 cm
Total length of meter rule = 100 cm
0 22cm 100cm
-------------------------Δ------------------------------------
↓ ↓
200g m₂
Apply principle of moment
(200 g)(22 cm - 0) = m₂(100 cm - 22 cm)
(200 g)(22 cm) = m₂(78 cm)
m₂ = (200 g)(22 cm) / (78 cm)
m₂ = 56.41 g
Therefore, the mass of the rule is 56.41 g
Answer:
<h2>
E = 2.8028*10⁻¹⁹ Joules</h2>
Explanation:
The minimum energy needed to eject electrons from a metal with a threshold frequency fo is expressed as E = hfo
h = planck's constant
fo = threshold frequency
Given the threshold frequency fo = 4.23×10¹⁴ s⁻¹
h = 6.626× 10⁻³⁴ m² kg / s
Substituting this value into the formula to get the energy E
E = 4.23×10¹⁴ * 6.626 × 10⁻³⁴
E = 28.028*10¹⁴⁻³⁴
E = 28.028*10⁻²⁰
E = 2.8028*10⁻¹⁹ Joules
