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rodikova
8 days ago
11

A mover hoists a 50 kg box from the ground to a height of 2 m. What was the change in the box's energy

Physics
1 answer:
kicyunya [2.9K]8 days ago
3 0

Response:

980 J

Clarification:

The energy change of the box is equivalent to its change in gravitational potential energy:

\Delta U = m g \Delta h

where

m = 50 kg signifies the mass of the box

g = 9.8 m/s^2 represents the acceleration caused by gravity

\Delta h= 2mindicates the height change of the box

By substituting the values, we discover

\Delta U = (50 kg)(9.8 m/s^2)(2 m)=980 J

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At a location where the acceleration due to gravity is 9.807 m/s2, the atmospheric pressure is 9.891 × 104 Pa. A barometer at th
serg [3228]

Answer:

8616.7468 \ kg/m^3

Explanation:

The measurement of pressure is indicated as p=\rho gh where p denotes the pressure, \rho signifies density, and h represents height

Given values include pressure p=9.891\times 10^4\ Pa, gravity's acceleration g=9.9870\ m/sec^2, and height =1.163 m

\rho =\frac{p}{gh}=\frac{9.891\times 10^4}{9.870\times 1.163}=8616.7468 \ kg/m^3

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19 days ago
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A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt2+qt, with p = 0.36
ValentinkaMS [3091]

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.

5 0
1 month ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Maru [2979]

Answer:

The final size is nearly the same as the initial size because the increase in size1.055\times 10^{- 7} is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass, m_{p} = 1.67\times 10^{- 27} kg

Initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This proton energy is \simeq 250 GeV

Thus, the speed of the proton, v\simeq c

The time to cover 1 km = 1000 m of distance is calculated as:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

According to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
20 days ago
A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at
Ostrovityanka [2820]

Answer

Given:

Wavelength = λ = 18.7 cm

                  = 0.187 m

Amplitude, A = 2.34 cm

Velocity, v = 0.38 m/s

A)  Calculate the angular frequency.

     f = \dfrac{v}{\lambda}

     f = \dfrac{0.38}{0.187}

     f =2.03\ Hz

Angular frequency,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) Calculate the wave number:

      K = \dfrac{2\pi}{\lambda}

     K= \dfrac{2\pi}{0.187}

    K =33.59\ m^{-1}

C)

Since the wave is traveling in the -x direction, the sign is positive between x and t

y (x, t) = A sin(k x - ω t)

y (x, t) = 2.34 sin(33.59 x - 12.75 t)

4 0
1 month ago
2. Turn off the Parallel line and turn on the Line through focal point. Move the light bulb around. What do you notice about the
serg [3228]

Answer:

The convergence of light rays redirects them toward the focal point, resulting in a magnifying effect.

Explanation:

8 0
24 days ago
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