A 10-kg sled carrying a 30-kg child glides on a horizontal, frictionless surface at a speed of 6.0 m/s toward the east. The chil
d jumps off the back of the sled, propelling it forward at 20 m/s. What was the child’s velocity in the horizontal direction relative to the ground at the instant she left the sled?
1 answer:
Answer:
- 1.33m/s
Explanation:
We choose the system to be the child and the sled. The surface is friction less, which means that there are no forces exerted on the system horizontally. This means that the horizontal momentum component of the system is constant and conserved.
So we can use the conservation momentum principle to find the velocity of the child just after he leaves the sled.
This is shown in the attached file.
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Answer:
t = 3.48 s
Explanation:
The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:
where,
v₀ = initial speed = 110 ft/s
Therefore,
<u>t = 3.48 s</u>
Answer:
2.00 m/s²
Explanation:
Given
The Mass of the metal safe, M = 108kg
Pushing force applied by the burglar, F = 534 N
Co-efficient of kinetic friction, = 0.3
Now,
The force against the kinetic friction is given as:
Where,
N = Normal reaction
g= acceleration due to the gravity
Substituting the values in the above equation, we get
or
Now, the net force on to the metal safe is
Substituting the values in the equation we get
or
also,
acceleration of the safe
Therefore, the acceleration of the metal safe will be
acceleration of the safe=
or
acceleration of the safe=
or
acceleration of the safe=
Hence, the acceleration of the metal safe will be 2.00 m/s²