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gtnhenbr [62]
3 years ago
8

The measurement 0.025563 g should be reported as?

Physics
1 answer:
Tomtit [17]3 years ago
7 0
The given mass is 0.025563 g.

Examine the given choices.
a. 0.026 g
This uses 2 significant digits, with rounding to the 3rd decimal place.

b. 2.5 x 10² g = 250 g.
It is incorrect.

c. 0.025 g.
This uses 2 significant digits. It is inaccurate because it does not use rounding to the 3rd decimal place.

d. 0.02 g
This uses one significant digit. It is incorrect for representing the given data.

Answer: a.  0.026 g
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Answer:

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Explanation:

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3 years ago
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Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at
Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

v = \dfrac{2\pi R}{T}

v = \dfrac{2\pi\times 8}{4.30}

  v = 11.69 m/s

now, Force does the ring push on her at the top

- N - m g = \dfrac{-mv^2}{R}

N + m g = \dfrac{mv^2}{R}

N = \dfrac{mv^2}{R}- m g

N = m(\dfrac{v^2}{R}- g)

N = 58\times (\dfrac{11.69^2}{8}- 9.8)

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0
3 years ago
A man stands on a platform that is rotating (without friction) with an angular speed of 1.0 rev/s; his arms are outstretched and
Airida [17]

Answer:

Explanation:

Given

N_1=1 rev/s

angular velocity \omega =2\pi N_1=6.284 rad/s

Combined moment of inertia of stool,student and bricks =6\ kg.m^2

Now student pull off his hands so as to increase its speed to suppose N_2 rev/s

\omega _2=2\pi N_2  

After Pulling off hands so final moment of inertia is

I_2=2\ kg-m^2

Conserving angular momentum  as no external torque is applied

I_1\omega _1=I_2\omega _2

6\times 6.284=2\times \omega _2

\omega _2=18.85\ rad/s

N_2=3 rev/s

7 0
3 years ago
Amotor supplied by 240V requires 12A to lift a 2000 lb at a rate of 25 ft/min, me power input to the motor is-
tensa zangetsu [6.8K]

Answer:

Power input, P = 2880 watts

Explanation:

It is given that,

Voltage of the motor, V = 240 V

Current required, I = 12 A

Weight lifted, W = 2000 lb

It is lifting at a speed of 25 ft/min. We need to find the power input to the motor. The product of current and voltage is called power input of the motor.

P=I\times V

P=12\ A\times 240\ V

P = 2880 watts

So, the power input of the motor is 2880 watts. Hence, this is the required solution.

4 0
4 years ago
Why does the closed top of a convertible bulge when the car is riding along a highway? Why does the closed top of a convertible
iren2701 [21]

Answer:

Option D

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Explanation:

When considering airflow over and around a surface, from Bernoulli's equation, air flow regions with higher velocity have a lower pressure, and regions with lower velocity have a higher pressure.

The air outside the convertible is moving faster than the air inside the convertible. This leads to a higher pressure zone just below the surface of the roof (inside the car) causing the roof of the convertible to bulge upwards

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