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sergejj [24]
3 years ago
8

A salt shaker sits 0.102 m from

Physics
1 answer:
Ivahew [28]3 years ago
6 0

The maximum speed is 0.55 m/s

Explanation:

For an object in uniform circular motion, the force of friction between the object and the ground provides the centripetal force required to keep the body in motion. Therefore we can write:

\mu_s mg = m\frac{v^2}{r}

where the term on the left is the frictional force and the term on the right is the centripetal force, and where

\mu_s is the coefficient of static friction

m is the mass of the body

g is the gravitational acceleration

v is the speed of the body

r is the radius of the circular path

In this problem, we have:

\mu_s = 0.307

r = 0.102 m

g=9.8 m/s^2

Substituting and re-arranging, we find the maximum speed v at which the salt shaker can rotate:

v=\sqrt{\mu gr}=\sqrt{(0.307)(9.8)(0.102)}=0.55 m/s

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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Convert to the fractional equivalent and reduce 21.12
nignag [31]

The decimal point is placed after two digits starting from the end. For each decimal place, we can write the number divided by 100.

21.12 can be written as \frac{2112}{100}.

Divide the numerator and denominator by 2:

\frac{2112}{100}= \frac{156}{50}

The numerator and denominator can be divided by 2 again:

\frac{78}{25}

There is no other common factor between numerator and denominator other than 1. Hence, it is the reduced form.




3 0
2 years ago
Read 2 more answers
A football punter accelerates a .55 kg football
Ronch [10]

Answer:

17.6 N

Explanation:

The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

F=\frac{\Delta p}{\Delta t}

where

\Delta p is the change in momentum of the football

\Delta t=0.25 s is the time elapsed

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.55 kg is the mass of the football

u = 0 is the initial  velocity (the ball starts from rest)

v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we  find the force exerted on the ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N

5 0
3 years ago
a) ¿En qué posición es mínima la magnitud de la fuerza sobre la masa de un sistema masa-resorte? 1) x 0, 2) x A o 3) x A. ¿Por q
Tatiana [17]

Answer:

a) the correct answer is 1 , b) x=0   F=0, a=0

x= 0.050    F= -7.5 N,  a= -15 m/s²

x= 0.150     F= 22.5 N,  a=- 45 m/s²

Explanation:

a) In a mass - spring system the force is given by the Hooke force,

          F = - k x

Analyzing this equation we see that the outside is proportional to the elongation from the equilibrium position, therefore the force is zero when the spring is in its equilibrium position

the correct answer is 1

b) we assume that the given values ​​are from the equilibrium position of the spring.

Let's calculate the force

x = 0

      F = 0

x = 0.050

      F = - 150 0.050

      F = - 7.50 N

x = 0.150

      F = - 150 0.150

      F = - 22.5 N

let's use Newton's second law to find the acceleration

      F = m a

      a = F / m

x = 0 m

      a = 0

x = 0.050 m

      a = -7.50 / 0.50

      a = - 15 m / s²

x = 0.150 m

      a = - 22.5 / 0.50

      a = - 45 m/s²

TRASLATE

a) En un sistema masa – resorte  la fuerza es dada por la fuerza de Hoke,  

          F= - k x

analizando esta ecuación vemos que la fuera es proporcional a la elongación desde la posición de equilibrio, por lo tanto la fuerza es cero cuando el resorte esta en su posición de equilibrio

la respuesta correcta es  1

b)suponemos que los valores dados son desde la posición de equilibrio del resorte.

Calculemos la fuerza  

x=0  

              F= 0

x=0.050  

              F = - 150 0.050

              F= - 7.50 N

x= 0.150  

                F= - 150 0.150

                F= - 22.5 N

usemos la segunda ley de Newton para encontrar la aceleración

          F = m a

          a = F/m

x =0  m

        a = 0

x= 0.050 m

         a = -7.50/ 0.50

          a =- 15 m/s²

x= 0.150 m

          a= - 22.5 / 0.50

          a= - 45 m/s²

7 0
3 years ago
How does changing the potential difference in a circuit affect the current and the resistance?
Effectus [21]

Answer:

The current will be increased and also for the resistance.

Explanation:

The analysis of a direct current circuit can give us the explanation we need. Using the ohm law, which tells us that the voltage is equal to the product of the current by the resistance we have:

V=I*R\\where\\V= voltage [V]\\I= amperes [amp]\\R=resistance [ohm]\\

The voltage is equal to the potential difference therefore we will have these expressions:

I=\frac{V}{R} \\R= \frac{V}{I}

If we increase the potential differential or circuit voltage, the current will also increase and so does the resistance by increasing the voltage. If we put numerical values in the equation given before, we can confirm this fact.

4 0
3 years ago
A railroad car with a mass of 30,000 kg is moving at 2.0 m/s when it runs into an at-rest freight car with an equal mass. The ca
11Alexandr11 [23.1K]
The total momentum of the system is preserved through the collision.

Note that momentum is
P = m*v
where m = mass
v = velocity.

Initial momentum:
P1 = (30000 kg)*(2 m/s) = 60000 (kg-m)/s for the moving car
P2 = 0 for the starionary car.

Final momentum:
P3 = (30000 + 30000)*v = 60000v (kg-m)/s

Because momentum is preserved,
P3 = P1 + P2
60000v = 60000
v = 1 m/s
The final velocity is 1 m/s.

Answer: 1.0 m/s
4 0
3 years ago
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