If a truck is travelling east on a straight road and travels 100 meters in 25s what is the truck's velocity?
1 answer:
Answer:
Velocity of the truck, v = 4 m/s
Explanation:
It is given that,
Distance covered by the truck, d = 100 m
Time taken by the truck, t = 25 s
Let v is the velocity of the truck. We know that the velocity is a vector quantity. Mathematically, it is given by :
v = 4 m/s
So, the velocity of the truck is 4 m/s. Hence, this is the required solution.
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The rod has a mass of m = 4.4 kg and a length of L = 0.4 m.
Its polar moment of inertia is
J = (mL²)/12
= (1/12) * [(4.4 kg)*(0.4 m)²]
= 0.05867 kg-m²
The mass of the bullet is 0.3 g.
If its velocity is v m/s, then its linear momentum is
P = (0.3 x 10⁻³ kg)*(v m/s)
Its linear momentum perpendicular to the rod is
P*sin(60°) = 2.5981 x 10⁻⁴ v (kg-m)/s
The angular momentum about the center of the rod when the bullet strikes is
T = (2.5981 x 10⁻⁴ v (kg-m)/s)*(0.2 m) = 5.1962 x 10⁻⁵ v (kg-m²)/s
Because the bullet lodges into the end of the rod, the combined polar moment of inertia is
J + (0.3 x 10⁻³ kg)*(0.2 m)² = 0.05867 + 1.2 x 10⁻⁵ = 0.0587 kg-m²
The initial angular velocity is ω = 17 rad/s.
Because angular momentum is conserved, therefore
5.1962 x 10⁻⁵ v (kg-m²)/s = (0.0587 kg-m²)*(17 rad/s)
v = 19204 m/s
Answer: 19204 m/s
Its polar moment of inertia is
J = (mL²)/12
= (1/12) * [(4.4 kg)*(0.4 m)²]
= 0.05867 kg-m²
The mass of the bullet is 0.3 g.
If its velocity is v m/s, then its linear momentum is
P = (0.3 x 10⁻³ kg)*(v m/s)
Its linear momentum perpendicular to the rod is
P*sin(60°) = 2.5981 x 10⁻⁴ v (kg-m)/s
The angular momentum about the center of the rod when the bullet strikes is
T = (2.5981 x 10⁻⁴ v (kg-m)/s)*(0.2 m) = 5.1962 x 10⁻⁵ v (kg-m²)/s
Because the bullet lodges into the end of the rod, the combined polar moment of inertia is
J + (0.3 x 10⁻³ kg)*(0.2 m)² = 0.05867 + 1.2 x 10⁻⁵ = 0.0587 kg-m²
The initial angular velocity is ω = 17 rad/s.
Because angular momentum is conserved, therefore
5.1962 x 10⁻⁵ v (kg-m²)/s = (0.0587 kg-m²)*(17 rad/s)
v = 19204 m/s
Answer: 19204 m/s