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Anni [7]
2 years ago
11

When the acceleration of a mass on a spring is zero, the velocity is at a

Physics
1 answer:
Sergeu [11.5K]2 years ago
6 0

1) Maximum

2) Maximum

Explanation:

The force acting on a mass on a spring is given by Hooke's law; in magnitude:

F=kx

where

F is the force

k is the spring constant

x is the displacement

Also we know from Newton's second law that we can write

F=ma

where

m is the mass

a is the acceleration

So we can write the equation as

ma=kx (1)

From this relationship, we see that the acceleration is directly proportional to the displacement.

On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by

E=\frac{1}{2}kx^2+\frac{1}{2}mv^2=const. (2)

where the first term is the elastic potential energy while the second term is the kinetic energy, and where

v is the velocity of the mass

From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.

Therefore this means that:

- When acceleration increases, velocity decreases

- When acceleration decreases, velocity increases

Therefore, the two answers here are:

- When the acceleration of a mass on a spring is zero, the velocity is at a  maximum

When the velocity of a mass on a spring is zero, the acceleration is at a  maximum

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Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 15.0 ∘ west of north,
Darina [25.2K]

Answer:

The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

Explanation:

The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.

The angle between the 2 forces and displacement is ∅ = 15°.

First we have to calculate the work done by the individual force and then we can calculate the total work.

The work done on a particle by a constant force F during a straight line displacement s is given by following formula:

W = F*s

W = F*s*cos∅

With ∅ = the angles between F and s

The magnitude of the force acting on the supertanker is F of tugboat1 = F of tugboat 2 = F = 2.2 * 10^6 N

The total work done can be calculated as followed:

Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2

Wtotal = 2Fs*cos∅

Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°

Wtotal = 3.44*10^9 Nm = <u>3.44 *10^9 J</u>

<u />

The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

5 0
2 years ago
CAN someone help ASAP?
lyudmila [28]

Answer:someone help me

Explanation:

7 0
2 years ago
What is the electrical force between q2 and q3? Recall that k = 8. 99 × 109 N•meters squared over Coulombs squared. 1. 0 × 1011
nlexa [21]

Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.

<h3>What is electric force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The electric force in the second case will be the same as in the first case. Therefore the force on the particle will be the same.

\rm F= K\frac{q_2q_3}{r^2}

\rm F= 9\times 10^9 \times \frac{1.6 \times 10^{-13}\times 1.6\times10^{-13}}{(0.5)^2}

\rm F=  - 1. 1 \times 10^{11 }N

Hence the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.

To learn more about the electric force refer to the link;

brainly.com/question/1076352

4 0
2 years ago
Question 10 (1 point)
torisob [31]

Answer:

It does not hit the students face because the speed of the balloon slows down as energy is lost through thermal.

Explanation:

3 0
2 years ago
A sculpture is suspended in equilibrium by two cables, one from a wall and the other
aleksandrvk [35]

Answer:

T_1=6655.295917 \approx 6655.3N

Explanation:

From the question we are told that

Angle of cable 2 \theta=37.0\textdegree

Weight of sculpture W=5000 N

Generally the Tension from cable 2 T_2 is mathematically given by

   T_2sin37\textdegree=5000N

   T_2=5000N/sin37\textdegree

   T_2=8308.2N

Generally the Tension from Cable 1 T_1 is mathematically given by

   T_1=T_2 cos37\textdegree

   T_1=8308.2* cos 37\textdegree

   T_1=6655.295917 \approx 6655.3N

7 0
3 years ago
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