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cupoosta [38]
3 years ago
11

A person is standing on a scale placed on the floor of an elevator. At time t1, the elevator is at rest and the reading on the s

cale is 500N. At a later time t2, the person is still standing on the scale and the reading on the scale is 400N . Based on the scale readings, which of the following statements about the motion of the elevator could be true at time t2?
a. The elevator is moving upward at constant speed.
b. The elevator is moving upward and slowing down.
c. The elevator is moving downward at constant speed.
d. The elevator is moving downward and slowing down.
e. The elevator is at rest.
Physics
2 answers:
sergiy2304 [10]3 years ago
7 0

the following statements about the motion of the elevator could be true at time t2 is

c. The elevator is moving downward at constant speed.

Explanation:

  • Under given situation, the elevator is at rest and the reading on the scale is 500N.
  • After some time t2, the person is still standing on the scale and the reading on the scale is 400N .
  • It is because if you stand on a scale in an elevator which is accelerating upward, your body will feel heavier because the elevator's floor pressure which presses harder on your feet, and this is why the scale will show a higher reading than the time when the elevator is at rest.
  • Similarly on the other hand, when the elevator accelerates downward, your body will feel lighter. The force which is exerted by the scale is called as the apparent weight; which means it does not change with constant speed.
  • Applying Newton's second law, which concludes about this particular statement about force exerted on a body when at rest and when in motion.

Gennadij [26K]3 years ago
7 0

Answer:

id say a

Explanation:

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Two ions with masses of 4.39×10−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic field is
Ilia_Sergeevich [38]

Answer:

7.2 cm

Explanation:

magnetic field, B = 0.301 T

speed, v = 7.92 x 10^5 m/s

mass, m = 4.39 x 10^-27 kg

q = 1.6 x 10^-19 C

The radius of singly changed ion is given by

r = \frac{mv}{Bq}

where, m is the mass of ion, v be the speed of ion, B is the magnetic field and q be the charge

r = \frac{4.39\times 10^{-27}\times 7.92 \times 10^{5}}{0.301\times 1.6\times 10^{-19}}

r = 0.072 m

r = 7.2 cm

5 0
3 years ago
What's the frequency of a wave with a wavelength of 10 and velocity of 200m/s?
uranmaximum [27]

Answer:

\boxed {\boxed {\sf 20 \ Hz}}

Explanation:

The frequency of a wave can be found using the following formula.

f=\frac{v}{\lambda}

where <em>f</em> is the frequency, <em>v</em> is the velocity/wave speed, and λ is the wavelength.

The wavelength is 10 meters and the velocity is 200 meters per second.

  • 1 m/s can also be written as 1 m*s^-1

Therefore:

v= 200 \ m*s^{-1} \\\lambda = 10 \ m

Substitute the values into the formula.

f=\frac{200 \ m*s^{-1}}{10 \ m}

Divide and note that the meters (m) will cancel each other out.

f=\frac{200 \ s^{-1}}{10 \ }

f=20 \ s^{-1}

  • 1 s^-1 is equal to Hertz
  • Therefore, our answer of 20 s^-1 is equal to 20 Hz

f= 20 \ Hz

The frequency of the wave is <u>20 Hertz</u>

7 0
3 years ago
A 9 newtwon charge is Felt on a (1x10^-4) C charge from another charge that is 3 meters away. What is the magnitude of that othe
Marina86 [1]

The magnitude of that other charge will be 9×10⁻⁵ C. The force on the charge is inverse of the distance.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The magnitude of that other charge is found as;

\rm F = \frac{Kq_1q_2}{r^2} \\\\ \rm 9  = \frac{9 \times 10^9 \times 1\times 10^{-4}\times q_2}{(3)^2} \\\\ q_2=9\times 10^{-5} \ C

Hence, the magnitude of that other charge will be 9×10⁻⁵ C.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

#SPJ1

6 0
2 years ago
How much work is done on an object that is moved to acquire a displacement of 5 meters when 500 Newtons of force was exerted?​
grandymaker [24]

Answer:

2,500 Joules (J) or Newton Meter (N M)

Explanation:

Work = Force x Distance

The force in this equation is 500 Newtons. The distance (displacement) is 5 meters. Plug it into the equation above.

Work = 5m x 500n

Work = 2,500 Joules or Newton-Meters.

Therefore 2,500 Joules or Newton Meters of work is done on an object.

3 0
3 years ago
c. A car was running with a velocity of 20m/s. what will be its velocity after 30s if it's acceleration is 5m/s2​
weqwewe [10]

\large \mathfrak{Solution : }

let's use first equation of motion to solve this ;

  • \boxed{ \boxed{ v = u + at}}

  • v = 20 + (5 \times 30)

  • v = 20 + 150

  • v = 170 \:  \: m/s

Velocity after 30 seconds = 170 m/s

4 0
3 years ago
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