A car is moving with a speed of 32.0 m/s. the driver sees an accident ahead and slams on the brakes, causing the car to slow dow

Explanation:

The formula for the electric field produced by an infinite sheet of charge is outlined below.

               E =

where,   is the surface charge density

Following this, the formula for the electric force acting on a proton is given as:

             F = eE

where,    e is the charge of a proton

According to Newton's second law of motion, the overall force on the proton can be expressed as follows.

                       F = ma

                 a =

                    =

                     =

According to kinematic equations, the proton's speed in the perpendicular direction can be described as follows.

                     =

                     =

                     = 683.974 m/s

Thus, the overall speed of the proton can be calculated as follows.

                v' =

                    =

                    = 1178.73 m/s

Consequently, we conclude that the proton's speed is 1178.73 m/s.

Answer:

Q = 12.5 kJ

Explanation:

The formula used to compute heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

ΔH° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

ΔH° = 35.0 kJ/mol

Using these two reactions, we can determine the heat change.

Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

2A + B + 2A2B -------> 2AB + A2

So, for the enthalpy, the values are summed:

H°3 = -25 + 35 = 10 kJ/mol

Now we can calculate the heat:

Q = 10 * 2.5 = 25 kJ

However, as we have 2A in the reaction, it does not maintain a 1:1 mole ratio, instead, it is 1:2, which requires us to adjust; thus:

Q = 25 / 2 = 12.5 kJ

Answer:

Statements 4, 6 & 7 are incorrect.

Explanation:

In any elastic collision, the overall momentum vector sum of the system remains zero.

In this scenario, an elastic collision occurs between the ball and a stationary wall. The ball's velocity will consistently revert after the impact, leading to a change in direction of momentum.

The initial momentum of the ball is represented as:

where:

m = mass of the ball

v = initial velocity of the body

post-collision for the elastic interaction:

• Here, the momentum changes solely in direction, thus contradicting statement 7.

• During the impact, both the ball and the wall exert forces on each other that are equal and opposite. The wall remains motionless, while the ball is influenced by the wall's reaction force, performing work on it, which contradicts statement 4.

• Given that this collision is elastic, the ball's form and dimensions do not alter.

• The previous points clearly indicate that not all provided statements hold true, thus violating statement 6.

Response:

0.60 m/s

Details:

The average speed between times t = a and t = b can be expressed as:

v_avg = (x(b) − x(a)) / (b − a)

Given the function x(t) = 0.36t² − 1.20t, and considering the interval from 1.0 to 4.0:

v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)

v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0

v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0

v_avg = [0.96 − (-0.84)] / 3.0

v_avg = 0.60

The average speed calculated is 0.60 m/s.