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BartSMP [9]
3 years ago
10

__________ ___________ are movements you do to loosen up your joints.

Physics
2 answers:
ss7ja [257]3 years ago
8 0

Answer:

Dynamic stretches

Explanation:

Dynamic stretches are active movements where joints and muscles go through a full range of motion. When you do dynamic stretches you do not hold the stretch, instead you stretch with movement.

Gala2k [10]3 years ago
4 0
The answer is B! Hoped o helped !have gray day I did Clase de AP – Unidad 2 Hoy es el 18 de noviembre
1.A.1: Identify the main idea.
2.A.1: Identify and/or describe content and connections among cultural topics. 4.A.2: Deduce the meaning of unfamiliar words or expressions
5.A.3: Provide and obtain relevant information in spoken exchanges.
Pregunta esencial:
1. ¿Como defines la belleza?
•
Calentamiento
asignacion en Teams)
el 19 de noviembre (Escribe en la
• ¿Cuál es la importancia de la moda en la vida social de las personas?
¿Cual es el impacto de la moda en la economia del individuo?
Tema de la unidad: La belleza y la estética
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A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 440 N
irinina [24]

Answer:

a) 0.275 m b) 13.6 J

Explanation:

In absence of friction, the energy is exchanged between the spring (potential energy) and the cookie (kinetic energy), so at any point, the sum of both energies must be the same:

E = ½ kx2 + ½ mv2

If we take as initial state, the instant when the cookie is passing through the spring’s equilibrium position, all the energy is kinetic, and we know that is equal to 20.0 J.

After sliding to the right, while is being acted on by a friction force, it came momentarily at rest. At this point, the initial kinetic energy, has become potential elastic energy, in part, and in thermal energy also, represented by the work done by the friction force.

So, for this state, we can say the following:

Ki = Uf + Eth = ½* k*d2 + Ff*d

20.0J = ½ *440 N/m* d2 + 11.0 *d, where d is the compressed length of the spring, which is equal to the distance travelled by the cookie before coming momentarily at rest.

We have a quadratic equation, that, after simplifying terms, can be solved as follows, applying the quadratic formula:

d = -0.05/2 +/- √0.090625 = -0.025 +/- 0.3 = 0.275 m (we take the positive root)

b) If we take as our new initial status the moment at which the spring is compressed, and the cookie is at rest, all the energy is potential:

E = Ui = 1/2 k d²

In this case, d is the same value that we got in a), i.e., 0.275 m (as the distance travelled by the cookie after going through the equilibrium point is the same length that the spring have been compressed).

E= 1/2 440 N/m . (0.275)m² = 16.6 J

When the cookie passes again through the equilibrium position, the energy will be in part kinetic, and in part, it will have become thermal energy again.

So, we can write the following equation:

Kf = Ui - Ff.d = 16.6 J - 11.0 (0.275) m = 16.6 J - 3.03 J = 13.6 J

3 0
3 years ago
If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular
erastova [34]

Answer;

D. The car would begin to move in the direction it was headed in a straight line.

Explanation;

-Centripetal force is any net force causing uniform circular motion. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration.

-The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

-Therefore,If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced then the car would begin to move in the direction it was headed in a straight line.

6 0
4 years ago
Read 2 more answers
An airport has runways only 198 m long. a small plane must reach a ground speed of 39 m/s before it can become airborne. what av
Anarel [89]
S: 198 m 
v=39 m/s 
u=0
t=? 
a=?

v²=u²+2as
(39)²=(0)²+2(a)(198) 
1521=396a
1521/396=a
3.84 m/s^2 = a 

Hope I helped :) 
8 0
4 years ago
......................
Ierofanga [76]
Don’t trust those links they usually pull up your IP
4 0
3 years ago
An automobile having a mass of 2000 kg deflects its suspension springs 0.02 m under static conditions. Determine the nafural fre
alexandr1967 [171]

Answer:Frequency = 3.525 Hertz

Explanation:In static equilibrium, kd =mg

Where k= effective spring constant of the spring.

mg= The weight of the car.

d= static deflection.

Therefore, w =SQRTg/d

w = SQRT 9.81/0.02

w= 22.15 rad/sec

Converting to Hertz unit for frequency

1 rad/s = 0.1591

22.15rad/s=?

22.15 × 0.1591= 3.525 hertz

7 0
3 years ago
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