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Marat540
1 month ago
15

A 1.0-kilogram ball is dropped from the roof of a building 40. meters tall. What is the approximate time of fall? [Neglect air r

esistance.]
Physics
2 answers:
ValentinkaMS [3.4K]1 month ago
6 0
Under a constant acceleration, the distance covered over time can be expressed as x=(gt^2)/2, with g representing acceleration and t signifying time. In this scenario, the only acceleration at play is from gravity, approximated at 9.81m/s^2. We're interested in a distance of x=40m. By substituting this into the equation and solving it, we have 40=(9.81*t^2)/2, leading to 80 = 9.81*t^2, which results in 8.1549 = t^2, thus yielding t = 2.856s.
Maru [3.3K]1 month ago
3 0
H = 40 m is the height from which the ball is released.
m = 1 kg is the ball's mass

Assuming g = 9.8 m/s² and disregarding air resistance.

The vertical velocity at the start is zero.
Let t be the flight duration, then
40 m = (1/2)*g*(t s)² = 0.5*9.8*t²
t² = 40/4.9 = 8.1633
t = 2.857 s

Result: 2.9 s (rounded to the nearest tenth)
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Read each scenario below. Then select the answer that best completes each sentence.
Ostrovityanka [3204]

Answer:

The power used by raul's microwave must match the power consumed by katrina's because both microwaves took different durations to accomplish the same heating task.

Explanation:

The power output from a car engine is equivalent to that of a bicycle since both perform the same amount of work over time. Both raul and katrina shared a frozen meal, heating each portion in different microwaves. Katrina's portion was warm in one minute, whereas raul's portion required two minutes. Therefore, the power utilized by raul's microwave aligns with that of katrina's, given that it took longer to achieve the same result.

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2 months ago
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A plane monochromatic radio wave (λ = 0.3 m) travels in vacuum along the positive x-axis, with a time-averaged intensity I = 45
serg [3582]
The magnetic field is calculated to be -6.137 × T. Explanation: Given the radio wave wavelength of λ = 0.3 m and an intensity of I = 45 W/m² at times t = 0 and t = 1.5 ns, we determine Bz at the origin. We use the intensity formula relating to the electric field, which incorporates the known intensity of 45, the speed of light c = 3 × m/s², and ∈o as 8.85 × C²/N.m², leading us to E = 184.15. Consequently, applying the equations, we find B = -6.137 × T at the z-axis.
8 0
2 months ago
A baseball is hit high into the upper bleachers of left field. Over its entire flight the work done by gravity and the work done
serg [3582]

Answer:

Explanation:

As the baseball ascends, gravitational forces as well as air resistance act downward, whereas the displacement is moving upward which results in an angle of 180° between the force and displacement. Therefore, the work done by both the gravitational force and air resistance is negative, confirming option (d) as accurate.

6 0
2 months ago
Laser beam with wavelength 632.8 nm is aimed perpendicularly at opaque screen with two identical slits on it, positioned horizon
Softa [3030]

Response:

1 slit width = 0.158 mm, slit separation = 0.633 mm, distance between diffraction maxima = 12.7 mm

Explanation:

This involves defining several terms:

λ, the wavelength of the beam

D, the distance from the plane to the slit

x, the distance between minima in the diffraction pattern (in single slit setups)

w, the fringe width for double slit setups

1. In a double slit experiment, the fringe width is also recognized as the distance between Maxima.

Thus, w=λD/d, leading to d=λD/w when rearranged.

So, d= (632.8 x 10^-9 x 1 x 10^3)/1

which equals d= 0.633mm.

For single slit diffraction, minima are defined by a*sinΘ= m*x

where a is the slit width and m is an integer.

For small angles, it simplifies to Θ= (x/D) = (λ/a), allowing us to solve for a:

a = λD/a

a = (632.8x10^-9 x 1)/4

yielding a= 0.158mm.

2. A grating with 20 slits/mm gives d= 1/20mm= 0.05x10^-3.

To find y (the distance between maxima), apply y= λL/d:

Finally, y= (632.8 x 10^-9 x 1)/0.05x10^-3, which results in y= 12.7mm.

6 0
2 months ago
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