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Paraphin [41]
3 years ago
12

Calcular el diámetro que debe tener en el embolo mayor un gato hidráulico, para obtener una fuerza de 3500N. Cuando el embolo me

nor tiene un diámetro de 16 cm y se le aplica una fuerza de 250N
Physics
1 answer:
Airida [17]3 years ago
5 0

Responder:

0.6 m

Explicación:

Responder:

0,15 m

Explicación:

Paso uno:

dado

Fuerza F = 250N

diámetro d1- 16 cm am = 0.16 m

A1 = πd1 ^ 2/4 = 3.142 * 0.16 ^ 2/4

= 0.02 m ^ 2

Carga L = 3500N

A2 =?

Segundo paso:

sabemos que la presión aplicada a un extremo para levantar una carga en el otro extremo se expresa matemáticamente como

F1 / A1 = F2 / A

sustituir

250 / 0.2 = 3500 / A

cruzar multiplicar

A2 = 0.02 * 3500/250

A2 = 70.3 / 250

A2 = 0.28

sabemos que área = πd ^ 2/4

0.28 = πd2 ^ 2/4

0.28 = 3.142 * d2 ^ 2/4

0.28 * 4 / 3.142 = d2 ^ 2

0.36 = d ^ 2

d = √0.36

d = 0.6 m

el diámetro debe ser de 0.6 m

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A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24
alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

I=0.8*m*r^{2}

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}

m=40kg

moment of inertia

M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}

Kinetic energy of the rotation motion

K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J

Kinetic energy translational

K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J

Total kinetic energy  

K=3317.79J+4147.2J\\K=7464.99J

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m

6 0
4 years ago
A runner accelerated to 4 m/s^2 for 20 seconds before winning the race.How far did he/she run?
creativ13 [48]

Answer:

s=800 m

Explanation:

Given that,

Acceleration of a runner, a = 4 m/s²

Time, t = 20 seconds

We need to find the distance covered by her. Initially, she was at rest. It means its initial velocity is equal to 0. So, using second equation of motion as follows :

s=ut+\dfrac{1}{2}at^2

Herre, u = 0

s=\dfrac{1}{2}at^2\\\\s=\dfrac{1}{2}\times 4\times (20)^2\\\\s=800\ m

So, she will cover a distance of 800 m.

8 0
4 years ago
A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv
sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

K_{rel}=(\gamma-1)mc^2

\dfrac{K_{rel}}{mc^2}=\gamma-1

Put the value into the formula

\gamma=\dfrac{105.7}{0.511}+1

\gamma=208

We need to calculate the electron’s velocity

Using formula of velocity

\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}

\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}

\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1

v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2

Put the value into the formula

v^2=\dfrac{1-(208)^2}{-208^2}\times c^2

v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}

v=0.9999 c\ m/s

Hence, The electron’s velocity is 0.9999 c m/s.

6 0
3 years ago
Which object has the most gravitational potential energy?
Kipish [7]

Answer: An 8 kg book at a height of 3 m has the most gravitational potential energy.

Explanation:

Gravitational potential energy is the product of mass of object, height of object and gravitational field.

So, formula to calculate gravitational potential energy is as follows.

U = mgh

where,

m = mass of object

g = gravitational field = 9.81 m/s^{2}

h = height of object

(A) m = 5 kg and h = 2m

Therefore, its gravitational potential energy is calculated as follows.

U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 2 m\\= 98.1 J    (1 J = kg m^{2}/s^{2})

(B) m = 8 kg and h = 2 m

Therefore, its gravitational potential energy is calculated as follows.

U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 2 m\\= 156.96 J    (1 J = kg m^{2}/s^{2})

(C) m = 8 kg and h = 3 m

Therefore, its gravitational potential energy is calculated as follows.

U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 3 m\\= 235.44 J    (1 J = kg m^{2}/s^{2})

(D) m = 5 kg and h = 3 m

Therefore, its gravitational potential energy is calculated as follows.

U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 3 m\\= 147.15 J    (1 J = kg m^{2}/s^{2})

Thus, we can conclude that an 8 kg book at a height of 3 m has the most gravitational potential energy.

3 0
3 years ago
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