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scoundrel [369]
2 years ago
9

A 84-kg man stands on a bathroom scale inside an elevator. the scale measures in units of newtons. (a) the elevator accelerates

upward from rest at a rate of 1.20 m/s2 for 1.50 s. what does the scale read during this 1.50 s interval
Physics
1 answer:
valentina_108 [34]2 years ago
3 0

Answer:

  94 kg

Explanation:

The mass registered by the scale is based on the assumption that the force applied is due entirely to gravity. If the force is greater, then the indicated mass will be greater.

__

<h3>how many g's</h3>

As a fraction of the acceleration of gravity, the elevator's acceleration is ...

  (1.2 m/s²)/(9.8 m/s²) ≈ 6/49

<h3>net force</h3>

The force required to produce a given acceleration is found by the formula ...

  F = ma . . . . . . . force on mass m to produce acceleration 'a'

When the man is stationary on the scale, the upward force it supplies is balanced by the downward force on the man due to gravity. The force and the mass are proportional, and the constant of proportionality (the acceleration due to gravity) is used to calibrate the scale. More force is thus translated to a higher mass reading.

Since the man's net acceleration is upward at the rate of 6/49×g, the total force applied by the scale is (1 +6/49) = 55/49 times as great as when the man is stationary. This greater force gets translated to a greater mass reading.

The force is equivalent to what would be required to support a stationary man with a mass of ...

  (84 kg)(55/49) = 94 2/7 kg

The scale would read about 94 kg during the upward acceleration period.

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A microwave oven operates with sinusoidal microwaves at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the
Degger [83]

Answer:

F = 2 × 10⁻³ N

Explanation:

Given:

frequency, f = 2400 MHz

Height, h = 25cm = 0.25 m

Area of the base, A = 30 cm x 30 cm = 900 cm² = 0.09 m²

Energy of the  microwave, E = 0.50 mJ = 0.5 x 10⁻³ J

Now, the time taken by the wave from top to the base, t = h/c

here, c is the speed of the light

thus,

t = 0.25/(3 x 10⁸) = 8.33 x 10⁻¹⁰ s

The radiation pressure P_r = Intensity/c

now, the intensity is given as:

I = Power/ area

also,

Power = Energy/ time = 0.5 x 10⁻³ J/8.33 x 10⁻¹⁰ s = 600000 W

thus,

I = 600000 W/ 0.09 m² = 6666666.6 W/m²

substituting the value in the formula for pressure due to radiation, we have

P_r = 6666666.6 W/m²/(3 x 10⁸)

also

pressure = Force/ area

thus,

Force/ area = 6666666.6 W/m²/(3 x 10⁸)

or

Force (F) = (6666666.6 W/m² × 0.09 m²)/(3 x 10⁸)

or

F = 2 × 10⁻³ N

6 0
3 years ago
Read 2 more answers
A telephone pole has three cables pulling as shown from above, with F⃗ 1=(300.0iˆ+500.0jˆ) , F⃗ 2=−200.0iˆ , and F⃗ 3=−800.0jˆ .
Ray Of Light [21]

A) Net force in component form: F=100.0i-300.0j

B) Magnitude of the net force: 316.2, direction: -71.6^{\circ}

Explanation:

A)

The three forces given in this problem are:

F_1=300i+500j

F_2=-200i

F_3=-800 j

The three forces are given in component form, where the components with unit vector i is the component along the x-direction, while the components with unit vector j is the component along the y-direction.

In order to find the net force in component form, we just need to add the components of the three forces along each direction. Therefore:

- Along the x-direction:

F_x = F_{1x}+F_{2x}+F_{3x}=300+(-200)+0=100

- Along the y-direction:

F_y=F_{1y}+F_{2y}+F_{3y}=500+0+(-800)=-300

So, the net force in component form is

F=100.0i-300.0j

B)

The magnitude of a vector F is given by Pythagorean's theorem:

|F|=\sqrt{F_x^2+F_y^2}

where in this problem,

F_x=100 is the x-component

F_y=-300 is the y-component

Substituting,

|F|=\sqrt{(100)^2+(-300)^2}=316.2

The direction instead is given by

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{-300}{100})=-71.6^{\circ}

where the negative sign means the direction is below the positive x-axis.

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

8 0
3 years ago
If two bodies weighing 10 kg and 20 kg respectively are standing at the
kvv77 [185]

Answer:

potential energy = mgh , where m is mass of body , g is acceleration due to gravity and h is given height

since both of the body are at same height above the ground , so the body having greater mass will have more potential energy

that is 20 kg body will have higher P.E

Explanation:

7 0
3 years ago
What is kelvin scale?
Darina [25.2K]
<span>a scale of temperature with absolute zero as zero, and the triple point of water as exactly 273.16 degrees.</span>
4 0
4 years ago
Read 2 more answers
Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
larisa86 [58]

Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • Mass of the larger disk = M_2\ =\ 1.60\ kg
  • Mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the larger disk = R_2\ =\ 5.00\ cm\ =\ 0.05\ m
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)

From the pulley,

\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)

From the equation (1) and (2),

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = R_2\ =\ 0.05\ m

Let a_2 be the acceleration of the block while connecting to the larger pulley.\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2

4 0
3 years ago
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