If a beaker of water is placed under a broiler so that the heating coil is above the beaker. It is observed that only the surfac

1 hour = 3,600 seconds
1 km = 1,000 meters

75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s

The mean speed during the deceleration is

                                   (1/2)(20-5/6 + 0) = 10-5/12 m/s.

Traveling at this average speed for 21 seconds,
the bus covers

                        (10-5/12) × (21) = 218.75 meters.

Answer:

The acceleration of the platform is - 1.8 m/s²

Explanation:

The net force on a body causes that body to accelerate in the direction of the resultant force applied.

Setting up the force equilibrium for the configuration:

ma = 800 - mg

100a = 800 - 100×9.8

100a = - 180

100a = - 180

a = - 1.8 m/s²

This indicates that the body is falling downward.

The average velocity of the sled can be expressed with the formula vavg = s/t. Hello! The calculation for average velocity involves determining the distance traveled over time. Thus, vavg = Δx/Δt, where vavg represents average velocity, Δx is the distance traversed, and Δt is the duration of time taken. We know both the distance (s) and the time (t) required for the sled to cover that distance, which allows us to compute the average velocity using the formula vavg = s/t. Wishing you a great day!

To address this issue, we apply the de Broglie equation written as:

λ = h/mv
where h equals 6.626×10⁻³⁴ J·s

Solving for m, we substitute for v, which is 46.9 m/s:
9.74 × 10⁻³⁵ m = 6.626×10⁻³⁴ J·s / (m)(46.9 m/s)
Thus, we find that m = 0.145kg

Answer:

The canyon measures 50.176 meters deep.

Explanation:

The student drops a rock from the rim of the canyon, requiring us to ascertain the depth of the canyon—name how far the ground is below the cliff.

The data we have:

Time = t = 3.2 s

Initial velocity = = 0 m/s

Gravitational acceleration = g = 9.8 m/s²

Height = h =?

According to the second equation of motion

Given the initial velocity is zero, the right-hand side of the equation simplifies to zero

h = (0.5)(9.8)(3.2)²

h = 50.176 m

This calculation indicates that the rock dropped a distance of 50.176 meters to reach the canyon's base. Thus, the canyon depth is 50.176 meters.