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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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A soft-drink bottler purchases glass bottles from a vendor. The bottles are required to have internal pressure strength of at le
maria [59]

Answer: K =24 psi

Explanation:

Given: Standard deviation =3psi

Internal pressure strength =157psi

Number of random bottle =n=64

K= 3 × square root of 64

K= 3×8=24 psi

If mean internal pressure K fall below K,

157-1.3=155.7psi

At 2%:

0.16×64 = 10.24

5 0
3 years ago
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

5 0
3 years ago
Monochromatic coherent light shines through a pair of slits. If the wavelength of the light is decreased, which of the following
Law Incorporation [45]

Answer:

he correct answers are a, b

Explanation:

In the two-slit interference phenomenon, the expression for interference is

          d sin θ= m λ                       constructive interference

          d sin θ = (m + ½) λ             destructive interference

in general this phenomenon occurs for small angles, for which we can write

           tanθ = y / L

           tan te = sin tea / cos tea = sin tea

           sin θ = y / La

un

derestimate the first two equations.

Let's do the calculation for constructive interference

         d y / L = m λ

the distance between maximum clos is and

         y = (me / d) λ

this is the position of each maximum, the distance between two consecutive maximums

         y₂-y₁ = (L   2/d) λ - (L 1 / d) λ₁          y₂ -y₁ = L / d λ

examining this equation if the wavelength decreases the value of y also decreases

the same calculation for destructive interference

         d y / L = (m + ½) κ

         y = [(m + ½) L / d] λ

again when it decreases the decrease the distance

the correct answers are a, b

7 0
3 years ago
A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that
kherson [118]

Answer:

= 7.88 × 10^-12 T

Explanation:

From the above question, we are told that:

Kinetic Energy of the proton is K. E = 10.0 MeV

Step 1

We convert 10.0 MeV to Joules

1 Mev = 1.602 × 10-13 Joules

10.0 MeV = 10.0 × 1.602 × 10^-13 Joules = 1.602 × 10^-12 J

Hence, the Kinectic energy of a proton = 1.602 × 10^-12 J

Step 2

Find the Speed of the Proton

The formula for Kinectic Energy =

K.E = 1/ 2 mv²

Where

m = mass of the proton

v = speed of the proton

K.E of the proton = 1.602 × 10^-12 J

Mass of the proton = 1.6726219 × 10^-27 kilograms

Speed of the proton = ?

1.602 × 10^-12J = 1/2 × 1.6726219 × 10^-27 × v²

1.602 × 10^-12J = 8.3631095 ×10^-28 × v²

v² = 1.602 × 10^-12/8.3631095 ×10^-28

v = √(1.602 × 10^-12/8.3631095 ×10^-28)

v = 43772331.227m/s

v = 4.3772331227 × 10^7m/s

Approximately = 4.4 × 10^7 m/s

Step 3

Find the Magnetic Field of that region of space

The formula for Magnetic Field =

B = m v / q r

We are told that the proton executes a circular orbit, hence,

mv = √2m(KE)

m = Mass of the proton = 1.6726219 × 10^-27 kg

K.E of the proton = 1.602 × 10^-12 J

v = speed of the proton = 4.4 × 10^7 m/s

q = Electric charge = 1.6 × 10^-19 C

r = radius of the orbit = 5.80Ã10^10 m

= 5.8 × 10^10m

Magnetic Field =

=√ (2 × 1.6726219 × 10^-27 kg × 1.602 × 10^-12 J) /( 1.6 × 10^-19 C × 5.80 × 10^10 m)

= 7.88 × 10^-12 T

The magnetic field in that region of space is approximately 7.88 × 10^-12 T

4 0
3 years ago
Which of the following is incorrect for nuclear forces
densk [106]

Answer:

c they obey inverse square law

4 0
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