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

Impulse can be expressed as the integral of F(t) dt from 0.012 s to 0.062 s

Since the function F(t) is unknown, an estimation method will be applied.

The integral represents the area under the curve.

The task suggests approximating this area as a triangle.

For this triangle, the base measures: 0.062 s - 0.012 s = 0.050 s.

The height corresponds to the maximum force of 35 N.

Consequently, the area computes to [1/2] * (0.05 s) * (35 N) = 0.875 N*s.

Final result: 0.875 N*s

Solution:

The kinetic energy of the clam at an elevation of 5.0 m is 5.19 J and the velocity of the clam at that height is 9.71 m/s.

Explanation:

Throughout its motion, mechanical energy remains constant. We understand that mechanical energy is the summation of potential energy and kinetic energy. Potential energy = , Kinetic energy = and Mechanical energy = Initial kinetic energy is zero. At a height of 9.8 m, the mechanical energy of the clam with a mass of 0.11 kg and g=9.81 is calculated as follows: 0.11×9.81×9.8 = 10.58 J.

Mechanical energy of the clam at a height of 5.0 m = = . Given that mechanical energy is conserved, we can state that the mechanical energy of the clam at a height of 9.8 m is equal to that at 5.0 m. The representation is as follows:

10.58 = 10.58 – 5.39 =  5.19 = the clam's kinetic energy measures 5.19 J.

Lastly, the speed of the clam at 5.0 m is computed; thus, 5.19 = 94.36 = = 9.71 m/s. The clam's speed is determined to be 9.71 m/s.

Answer:

3000 kg.m/s

Explanation:

Momentum, often denoted as p, is the product of mass and velocity defined by p=mv where m is mass and v is velocity.

The change in momentum is described by where f and i denote final and initial states respectively. As the lorry comes to a halt, its final speed is zero. By substituting the provided values, we find that

Change in momentum= 6000(0-0.5)=-3000 kg.m/s

Response:

a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0

Clarification:

a)

• Electrostatic potential V is defined as the work achieved per unit charge, as conducted by the electrostatic force, which moves a distance d from infinity (considered the reference zero level).
• For a point charge, it can be represented mathematically:

• Since electrostatic force behaves linearly concerning charge, we can apply the superposition principle.
• This principle states that the cumulative potential at any given point is simply the sum of the individual potentials contributed by various charges, as if the others were absent.
• In our specific configuration, due to symmetry, the potential at each corner of the triangle is simply double that of the potential resulting from any charge at another corner, as demonstrated:

• The potential at point C registers as 8.99*10³ V.

b)

• The energy needed to move a positive charge of 5μC from infinity to point C is calculated by multiplying the potential at that point by the charge, explained below:

• The work required amounts to 0.045 J.

c)

• If we substitute one of the charges at point C with one of the opposite charge of equal magnitude, the following equation emerges:

• This indicates that the potential arising from both charges results in 0 at point C.

d)

• With point C's potential calculated as 0 and assuming V=0 at infinity too, we derive that bringing the charge of 5μC from infinity to point C requires no work, as there is no potential difference between the two locations.