Answer:
Kf > Ka = Kb > Kc > Kd > Ke
Explanation:
We can apply
E₀ = E₁
where
E₀: Mechanical energy at the beginning of the motion (top of the incline)
E₁: Mechanical energy at the end (bottom of the incline)
then
K₀ + U₀ = K₁ + U₁
If v₀ = 0 ⇒ K₀
and h₁ = 0 ⇒ U₁ = 0
we get
U₀ = K₁
U₀ = m*g*h₀ = K₁
we apply the same equation in each case
a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J
d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J
e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J
f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J
finally, we can say that
Kf > Ka = Kb > Kc > Kd > Ke
The words best describe the archer's results are The correct option is A.
<h3>What is accuracy?</h3>
When in an experiment, a value is measured 5 times, then if the values measured are same for most of the time or like three times out of five, it said to be accurate. The phenomenon is accuracy.
Precision is about comparing the values to each other then find them near to each other.
Given is an archer's targets on the target board.
Accuracy compares the experimental value to the theoretical value (desired target)
So, when all the values are close to each other but not nearest to the theoretical value, then it is said to be precise but not accurate.
Thus, the correct option is A.
Learn more about accuracy.
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Answer:
M L1 = m L2 torques must be zero around the fulcrum
M = m L2 / L1 = .3 kg * 28 cm / 22 cm = .382 kg
Answer:
1.74 m/s
Explanation:
From the question, we are given that the mass of the an object, m1= 2.7 kilogram(kg) and the mass of the can,m(can) is 0.72 Kilogram (kg). The velocity of the mass of an object(m1) , V1 is 1.1 metre per seconds(m/s) and the velocity of the mass of can[m(can)], V(can) is unknown- this is what we are to find.
Therefore, using the formula below, we can calculate the speed of the can, V(can);
===> Mass of object,m1 × velocity of object, V1 = mass of the can[m(can)] × velocity is of the can[V(can)].----------------------------------------------------(1).
Since the question says the collision was elastic, we use the formula below
Slotting in the given values into the equation (1) above, we have;
1/2×M1×V^2(initial velocity of the first object) + 1/2 ×M(can)×V^2(final velocy of the first object)= 1/2 × M1 × V^2 m( initial velocity of the first object).
Therefore, final velocity of the can= 2M1V1/M1+M2.
==> 2×2.7×1.1/ 2.7 + 0.72.
The velocity of the can after collision = 1.74 m/s
I think it is (c) but I’m not 100% sure
