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Zinaida [17]
2 years ago
12

A person who normally weighs 580 N is riding in an elevator that is moving upward, but slowing down at a steady rate. If this pe

rson is standing on a bathroom scale inside the elevator, what would the scale read
Physics
1 answer:
Zanzabum2 years ago
5 0

Answer:

M au = Fs - M g       au = upwards acceleration; Fs = scale reading

Fs = M (au + g)    scalar quantities where g is positive downwards and au is positive upwards - Fs is the net force acting on the person

If the acceleration is zero Fs = M g  and the scale reads the persons weight

If the elevator is decelerating then au is negative and the scale reading     Fs = (g - au) M     and the scale reading is less than the weight of the person

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Here stress is parallel to the surface of the body. So it's a Shear stress.
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3 years ago
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(b) A car of mass 3000 kg travels at a constant velocity of 5.0 m/s.
Tamiku [17]

Answer:

16.7 s

Explanation:

T= <u>Vf - Vo</u>          a= <u>F</u>

         a                    m

4,500 / 3000 = 1.5 (a)

30 - 5 / 1.5(a) = 16.7 s      

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3 years ago
What type of motion occurs when an object spins around an axis without altering its linear position?
dimulka [17.4K]
That would be only rotational motion
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3 years ago
In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so
Pepsi [2]

Answer:

(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

a_r=\omega^2R

Plug in the given values and solve for a_r. This gives,

a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2

Now, tangential acceleration is given by the formula:

a_t=R\alpha

Plug in the given values and solve for a_t. This gives,

a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

a_{Total}=\sqrt{(a_r)^2+(a_t)^2}

Plug in the given values and solve for total acceleration, a_{Total}. This gives,

a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

4 0
3 years ago
A 250 g block of ice is removed from the refrigerator at -8.0°C. How much thermal energy does the ice absorb as it warms to room
zlopas [31]

Answer:

Q = 114895 J

Explanation:

To find the thermal energy gained by the ice you use the following formula:

Q=mc(T_2-T_1)+H_f\ m

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T2: final temperature = 22°C

T1: initial temperature = -8.0°C

Hf: heat of fusion of water = 3.34*10^5 J/kg

c: specific heat of water = 4186 J/kg

By replacing the values of the parameters you have:

Q=(0.250kg)(4186J/kg\°C)(22+8)\°C+(3.34*10^5 J/kg)(0.250kg)\\\\Q=114895\ J

where you have considered that ice melts completely

3 0
3 years ago
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