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3 years ago

15

a passenger elevator operates at an average of 8 m/s if the 60th floor is 219 m above the first floor how long does it take the

elevator to go from the 1st floor to the 60th floor ?

1 answer:

MAVERICK [17]3 years ago

4 0

Answer:

The universal sign for choking is __________.

A.

two balled fists pressing the abdomen

B.

pointing at an open mouth

C.

two hands grasping the neck

D.

pretending to cough

Explanation:

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Block A, with mass mA, is initially at rest on a horizontal floor. Block B, with mass mB, is initially at rest on the horizontal

nexus9112 [7]

Answer:

(E) μs(mA +mB)g

Explanation:

We can apply for mB:

∑ Fx = mB*a (→)

⇒ Ffriction = mB*a ⇒ a = Ffriction / mB = μs*N / mB

⇒ a = μs*(mB*g) / mB ⇒ a = μs*g (acceleration of the system)

Now, for mA we have

∑ Fx = mA*a (→)

F - Ffriction = mA*a ⇒ F = mA*a + Ffriction

⇒ F = mA*(μs*g) + μs*(mB*g) ⇒ F = μs*g*(mA + mB)

We must know that the friction acts only between the two blocks

8 0

3 years ago

Does static electricity attract dead skin cells

lesantik [10]

Answer:

no i dont think so

Explanation:

6 0

3 years ago

A motor has a rotor (with armature) of moment of inertia ????m . The rotor is attached to a gear box of gear ratio G > 1. The

Inessa [10]

Answer:

hello your question is incomplete attached below is the complete question

answer : The moment of inertial felt by someone ( J ) is greater that the moment of inertia felt by the motor i.e. J > Jm

Explanation:

Gear ratio G > 1

a) Determine the moment of inertia felt by the motor

moment of inertia felt by Motor = moment of Inertia at the armature

b) Determine the moment of inertial felt by someone who is rotating the mass by hand

moment of inertia felt by someone is = J

The moment of inertial felt by someone ( J ) is greater that the moment of inertia felt by the motor

attached below is a detailed solution

3 0

3 years ago

What is the highest pHoney bees beat their wings, making a buzzing sound at a frequency of 2.3 × 102 hertz. What is the period o

VMariaS [17]

The period of the wave is 4.35 ms. The sound waves are called longitudinal waves

Explanation:

The period of a wave is related to its frequency by the equation:

$T=\frac{1}{f}$

where

T is the period

f is the frequency

For the bee in this problem, the frequency of the sound wave emitted by it is

$f=2.3 \cdot 10^2 Hz = 230 Hz$

Therefore, the period of the sound wave is

$T=\frac{1}{230}=4.35\cdot 10^{-3}s = 4.35 ms$

The sound wave is a type of wave called longitudinal wave. In longitudinal waves, the oscillation of the medium occurs in a direction parallel to the direction of motion of the wave: therefore in a sound wave, the particle of the medium (air, in this case) oscillate back and forth along the direction of propagation of the wave, forming alternating areas of higher density of particles (called compressions) and of lower density of particle (called rarefactions).

The other type of wave, instead, is called transverse wave. In a transverse wave, the oscillation of the wave occurs in a direction perpendicular to the direction of motion of the wave. An example of transverse waves are the electromagnetic waves, which consists of electric field and magnetic fields that vibrate in a plane perpendicular to the direction of motion of the wave itself.

Learn more about waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

5 0

3 years ago

A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr

Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration $a=g=9.8 m/s^2$ directed downward, initial vertical position $d=750 m$ , and initial vertical velocity $v_0 = 0$ . We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

$v_f^2 -v_i^2 =2ad$

substituting, we find

$v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s$

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

$v_f=\frac{121.2 m/s}{2}=60.6 m/s$

In order to do that, we use again the same SUVAT equation substituting $v_f$ with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

$v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m$

Which means that the heigth of the packet was

$h=750 m-187.4 m=562.6 m$

3 0

3 years ago