The speed of the pin after the elastic collision is 9 m/s east.
<h3>
Final speed of the pin</h3>
The final speed of the pin is calculated by applying the principle of conservation of linear momentum as follows;
m1u1 + mu2 = m1v1 + m2v2
where;
- m is the mass of the objects
- u is the initial speed of the objects
- v is the final speed of the objects
4(1.4) + 0.4(0) = 4(0.5) + 0.4v2
5.6 = 2 + 0.4v2
5.6 - 2 = 0.4v2
3.6 = 0.4v2
v2 = 3.6/0.4
v2 = 9 m/s
Thus, The speed of the pin after the elastic collision is 9 m/s east.
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Answer:
0.4
Explanation:
F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.
Since the mass is moving with a constant velocity, there's no acceleration hence
where N is the weight of object and \mu is coefficient of kinetic friction.
the subject

Substituting F for 8 N and N for 20 N

Therefore, coefficient of kinetic friction is 0.4
Answer:
138.3 days
Explanation:
Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.
The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.
T^2 is proportional to r^3
That is,
T^2/r^3 = constant
98^2 / 62^3 = T^2 / 78^3
Make T^2 the subject of formula.
T^2 = 98^2 / 62^3 × 78^3
T^2 = 19123.2
T = sqrt ( 19123.2 )
T = 138.2867 days
Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.
Answer:
Thus, the change in the weight of the person is 1.6N , option c is correct.
Explanation:
(a) The momentum of the proton is determined as 5.17 x 10⁻¹⁸ kgm/s.
(b) The speed of the proton is determined as 3.1 x 10⁹ m/s.
<h3>
Momentum of the proton</h3>
The momentum of the proton is calculated as follows;
K.E = ¹/₂mv²
where;
- m is mass of proton = 1.67 x 10⁻²⁷ kg
- v is speed of the proton = ?
<h3>Speed of the proton</h3>
v² = 2K.E/m
v² = (2 x 50 x 10⁹ x 1.602 x 10⁻¹⁹ J)/(1.67 x 10⁻²⁷)
v² = 9.6 x 10¹⁸
v = 3.1 x 10⁹ m/s
<h3>Momentum of the proton</h3>
P = mv = (1.67 x10⁻²⁷ x 3.1 x 10⁹) = 5.17 x 10⁻¹⁸ kgm/s
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