1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Anni [7]
1 year ago
11

An object, whose mass is 0.660 kg, is attached to a spring with a force constant of 132 N/m. The object rests upon a frictionles

s, horizontal surface (shown in the figure below).
An object labeled m is attached to the right end of a horizontal spring, and the left end of the spring is attached to a wall. The spring is stretched horizontally such that the object is displaced by a distance A to the right of its equilibrium position.
The object is pulled to the right a distance A = 0.120 m from its equilibrium position (the vertical dashed line) and held motionless. The object is then released from rest.
(a)
At the instant of release, what is the magnitude of the spring force (in N) acting upon the object?
N
(b)
At that very instant, what is the magnitude of the object's acceleration (in m/s2)?
m/s2
(c)
In what direction does the acceleration vector point at the instant of release?
Toward the equilibrium position (i.e., to the left in the figure).
Away from the equilibrium position (i.e., to the right in the figure).
The direction is not defined (i.e., the acceleration is zero).
You cannot tell without more information.

Physics
1 answer:
Arada [10]1 year ago
7 0

Answer:

See below

Explanation:

a)  Spring force at release =  k * d = 132 N/m * .120 m = 15.84 N

b) F = ma

   15.84 = (.660 kg)(a)     a = 24 m/s^2

c) Toward the left ....the object is accelerated to the left

You might be interested in
An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4
Maslowich

Answer:

Average speed = 10,000 m/s

Explanation:

Given the following data;

Distance = 2m

Time = 0.0002secs

To find the average speed;

Average speed = distance/time

Average speed = 2/0.0002

Average speed = 10,000 m/s

Therefore, the average speed of the

electron is 10,000 meters per seconds.

7 0
3 years ago
You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?
In-s [12.5K]
-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is  14 cos(</span><span>51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.


As usual when we're discussing this stuff, we completely ignore air resistance.
</span>
4 0
3 years ago
Read 2 more answers
The significant feature of a Cepheid variable is that there is a relationship between two intrinsic parameters, one of which can
Ghella [55]

Answer:

Period of brightness variation and luminosity.

Explanation:

The Cepheid variables are used as distance indicators. This requires estimation of periods and (usually) intensity-mean magnitudes in order to establish a period—apparent luminosity relation. It is particularly important for the techniques employed to be as accurate and efficient as possible.

5 0
2 years ago
NEED HELP ASAP
Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:  

p_{o}=p_{f} (1)  

Where:  

p_{o}=m_{1} V_{o} + m_{2} U_{o} (2)  

p_{f}=(m_{1} + m_{2}) V_{f} (3)

m_{1}=110 kg is the mass of the first football player

V{o}=-7 m/s is the velocity of the first football player (to the south)

m_{2}=75 kg  is the mass of the second football player

U_{o}=4 m/s is the velocity of the second football player (to the north)

V_{f} is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f} (4)  

Isolating V_{f}:

V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}} (5)

V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg} (6)

V_{f}=-2.54 m/s (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy \Delta K is defined as:

\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2} (8)

Simplifying:

\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2}) (9)

\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2}) (10)

Finally:

\Delta K=-2351.25 J (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0
3 years ago
Do you even juul bro?
blsea [12.9K]

Answer:

No

Explanation:

5 0
2 years ago
Read 2 more answers
Other questions:
  • Barry is conducting an experiment and rolls a tennis ball down a ramp. Which best describes the motion of the tennis ball? It do
    9·2 answers
  • Which free body diagram is in equilibrium? Question 5 options:
    12·1 answer
  • A doctor counts 68 heartbeats in 1.0 minute. What are the corresponding period and frequency of the heart rhythm
    12·1 answer
  • Is this right? if not how do u do it?
    8·1 answer
  • 9. Calculate the distance (in km) that Charlie runs if he maintains the average
    12·1 answer
  • I also need the direction Thank you!
    8·1 answer
  • It's possible for a determined group of people to pull an aircraft. Drag is negligible at low speeds, and the only force impedin
    6·1 answer
  • In an uncomformity between two layers of rock, how is the lower layer usally described
    11·1 answer
  • Three dogs (Spot, Fido, and Steinberg) are pulling on a chew toy. The chew toy is experiencing no acceleration. Spot is pulling
    8·1 answer
  • B)
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!