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postnew [5]
3 years ago
13

Now the elevator is moving downward with a velocity of v = -2.8 m/s but accelerating upward with an acceleration of a = 5.5 m/s2

. (Note: an upward acceleration when the elevator is moving down means the elevator is slowing down.) What is the force the bottom spring exerts on the bottom mass?
Physics
1 answer:
borishaifa [10]3 years ago
5 0

Answer:

160.75 N

Explanation:

The downward velocity has no effect on the force situation, it is only changes in velocity (plus, of course, gravity, which is always there) that require a force. At constant velocity, the bottom spring s_3 is supporting its mass m_3 to balance gravity.

As the elevator slows, though, it also ends up slowing down the spring arrangement, too. However, because the stretching takes time, it means that some damped harmonic motion will be set up in the spring chain.

When the motion has finally damped out, the net force the bottom spring s3 exerts on m3 has two components--that of gravity and of the deceleration of the elevator:

F_3net = m3 * (g + a) = 10.5×(9.81+5.5)= 10.5×15.31= 160.75 N

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State the law of conservation of linear momentum using Newton's third law of deduce this ​
MArishka [77]

Answer:

Derivation of Conservation of Momentum

Applying Newton's third law, these two impulsive forces are equal and opposite i.e. is equal to the change in momentum of the first object. is equal to the change in momentum of the second object. This relation suggests that momentum is conserved during the collision.

Explanation:

Hope it helps!!!

7 0
2 years ago
Consider two massless springs connected in series. Spring 1 has a spring constant k1, and spring 2 has a spring constant k2. A c
Andru [333]

Answer:

a. k = (1/k₁ + 1/k₂)⁻¹ b. k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

Explanation:

Since only one force F acts, the force on spring with spring constant k₁ is F = k₁x₁ where x₁ is its extension

the force on spring with spring constant k₂ is F = k₂x₂ where x₁ is its extension

Let F = kx be the force on the equivalent spring with spring constant k and extension x.

The total extension , x = x₁ + x₂

x = F/k = F/k₁ + F/k₂

1/k = 1/k₁ + 1/k₂

k = (1/k₁ + 1/k₂)⁻¹

B

The force on spring with spring constant k₃ is F = k₃x₃ where x₃ is its extension

Let F = kx be the force on the equivalent spring with spring constant k and extension x.

The total extension , x = x₁ + x₂ + x₃

x = F/k = F/k₁ + F/k₂ + F/k₃

1/k = 1/k₁ + 1/k₂ + 1/k₃

k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

8 0
3 years ago
Read 2 more answers
Temperature and pressure can change the what of a solute
igomit [66]
Temperature and pressure can change the solubility of a solute.
4 0
3 years ago
Which of the following statements is true? (A) The measured value for the pressure of a gas in a container is almost independent
Lana71 [14]

Answer:

The correct answer to the question is

Both A and B are true

Explanation:

The particles of a gas are free to move to occupy the entire volume in which they are placed due to the smallerinter molecular forces holding them together hence due to the face that pressure is a measure of the Force per unit area that is Pressure P = ( Force F)/ (Area A) then the force per unit area, exerted on the all of the container by the gaseous particles which are colliding with each other and with the walss of the container is fairly constant through out the surface oof the container

In the case of the liquid which are held on together by more stronger forces, the force per nit area exerted by the liquid particle is transmitted from one particle to the next until it reaches the container's surface. Then remembering that the force of gravity on the liquid is acting in one direction (that is downwards) the sum of the fprce due to the weight incrreases as we progress deaper into the liquid hence the pressure increases per unit depth

5 0
3 years ago
A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed
AleksandrR [38]

(a) 3.5 Hz

The angular frequency in a spring-mass system is given by

\omega=\sqrt{\frac{k}{m}}

where

k is the spring constant

m is the mass

Here in this problem we have

k = 160 N/m

m = 0.340 kg

So the angular frequency is

\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s

And the frequency of the motion instead is given by:

f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz

(b) 0.021 m

The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at

x = A

where A is the amplitude of the motion.

The maximum displacement is given by Hook's law:

F=kA

where

F is the force applied initially to the spring, so it is equal to the weight of the block:

F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N

k = 160 N/m is the spring constant

Solving for A, we find

A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m

3 0
3 years ago
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