Which type of nuclear decay emits two protons and two neutrons?
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The concept of this question can be well understood by listing out the parameters given.
- The mass of the block = 51 g = 51 × 10⁻³ kg
- The distance of the block from the table = 30 cm
- Length of the spring = 30 cm
The purpose is to determine the spring constant.
Let us assume that the two blocks are Block A and Block B.
At point A on block A, the initial velocity on the block is zero
i.e. u = 0
We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.
From the above formula,
The distance (S) = 30 cm; we need to convert the unit to meter (m).
- Since 1 cm = 0.01 m
- Then, 30cm = 0.3 m
The acceleration (g) due to gravity = 9.8 m/s²
∴
inputting the values into the equation above, we have;
By dividing both sides by 4.9, we have:
However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.
By applying the equation of the time period of a simple harmonic motion.
where the relation between time (t) and period (T) is:
T = 2t
T = 2(0.247)
T = 0.494 seconds
By making the spring constant k the subject of the formula:
Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.
Learn more about simple harmonic motion here:
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The given vectors quantities can be described by their properties of both
magnitude and direction.
- a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.
- b. The velocity vector extending from (0, 0) to (108.76, 50.714) on the coordinate plane is attached.
- c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.
Reasons:
a. The magnitude of the vector = 190 N
The direction in which the vector acts = East
Therefore, in vector form, we have;
= 190 × cos(0)·i + 190 × sin(0)·j = 190·i
The vector can be represented by an horizontal line, 190 units long
Coordinate points on the vector = (0, 0) and (190, 0)
The drawing of the vector with the above points using MS Excel is attached.
b. Magnitude of the velocity vector = 120 km/hr. 25° North of east
Solution;
The vector form of the velocity is; = 120 × cos(25)·i + 120×sin(25)·j, which gives;
= 120 × cos(25)·i + 120×sin(25)·j ≈ 108.76·i + 50.714·j
≈ 108.76·i + 50.714·j
Therefore, points that define the vector are; (0, 0) and (108.76, 50.714)
The drawing of the vector is attached
c. The magnitude of the vector = 60 m
The direction of the vector is southwest = West 45° south
The vector form of the displacement is = 60 × cos(45°)·i + 60 × sin(45°)·j
Which gives;
= 30·√2·i + 30·√2·j
Points on the vector are therefore; (0, 0), and (30·√2, 30·√2)
The drawing of the vector is attached
Learn more about vectors here:
