Answer:
Uniform acceleration means that the object is accelerating at a constant rate. Since there is still acceleration, velocity is not constant. In fact, the velocity is increasing. This means that the object is moving faster and faster at a constant rate.
This graph is a displacement-time graph so its gradient would represent the velocity of the object. The velocity increases hence the gradient should increase (which means that the graph must be steeper and steeper with increasing time).
Constant gradient of a displacement-time graph would mean that there is no acceleration since gradient of the graph does not change so velocity is constant.
Note that it is the<em> </em><em>rate</em><em> </em><em>of</em><em> </em><em>change</em> of velocity that is constant during uniform acceleration, not velocity.
Answer:
100
Explanation:
= Resistivity of axon
r = Radius of axon
t = Thickness of the membrane
= Resistivity of the axoplasm
Speed of pulse is given by
So, radius is given by
If radius is increased by a factor of 10 new radius will be
So, The radius will increase by a factor of 100.
Answer:
density
Explanation:
mecurys is significantly smaller and less massive than earth
A. The angle at which the arrow must be released to hit the bull's-eye is 20.7 °
B. The arrow will go over the branch.
<h3>A. How to determine the angle</h3>
- Range (R) = 74 m
- Initial velocity (u) = 33 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Angle (θ) = ?
R = u²Sine(2θ) / g
74 = 33² × Sine (2θ) / 9.8
Cross multiply
74 × 9.8 = 33² × Sine (2θ)
725.2 = 1098 × Sine (2θ)
Divide both sides by 1098
Sine (2θ) = 725.2 / 1098
Sine (2θ) = 0.6605
Take the inverse of sine
2θ = Sine⁻¹ 0.6605
2θ = 41.3
Divide both sides by 2
θ = 41.3 / 2
θ = 20.7 °
<h3>B. How to determine if the arrow will go over or under the branch</h3>
To determine if the arrow will go over or under the branch situated mid way, we shall determine the maximum height attained by the arrow. This can be obtained as follow:
- Initial velocity (u) = 33 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Angle (θ) = 20.7 °
- Maximum height (H) = ?
H = u²Sine²θ / 2g
H = [33² × (Sine 20.7)²] / (2 ×9.8)
H = 6.94 m
Thus, the maximum height attained by the arrow is 6.94 m which is greater than the height of the branch (i.e 3.50 m).
Therefore, we can conclude that the arrow will go over the branch
Learn more about projectile motion:
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In the first shell there can be a maximum of 2 electrons, then 8o in the second one and 8 in the third one.
