It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.
Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.
The missing part of the incomplete question is given below:
Which important step of scientific design is Shameka conducting?
repetition
replication
verification of results
using controlled variables
Answer:
Verification of results
Explanation:
The way toward gathering five examples of water from various sources is conveyed to confirm the outcome. By gathering water from five distinct areas of a similar source the analyst can genuinely find out the nature of the water in her region of remain.
On the off chance that after examples are tried it is found the water isn't sound, the outcomes would be acknowledged as it has been appropriately checked and a proper move would be made.
Thus, the correct answer is - verification of results
Answer:
v₁f = 0.5714 m/s (→)
v₂f = 2.5714 m/s (→)
e = 1
It was a perfectly elastic collision.
Explanation:
m₁ = m
m₂ = 6m₁ = 6m
v₁i = 4 m/s
v₂i = 2 m/s
v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i
v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)
v₁f = 0.5714 m/s (→)
v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i
v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)
v₂f = 2.5714 m/s (→)
e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1
It was a perfectly elastic collision.
Answer:
The actual angle is 30°
Explanation:
<h2>Equation of projectile:</h2><h2>y axis:</h2>
the velocity is Zero when the projectile reach in the maximum altitude:
When the time is vo/g the projectile are in the middle of the range.
<h2>x axis:</h2>
R=Range
**sin(2A)=2sin(A)cos(A)
<h2>The maximum range occurs when A=45°
(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>
Let B the actual angle of projectile
2B=60°
B=30°
Yes it is possible. Spectrum of emitted light depends upon the chemical composition of the source. and the way of its excitation. a clear example to us is that of sun.