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schepotkina [342]
2 years ago
7

if you apply a Force of F1 to area A1 on one side of a hydraulic jack, and the second side of the jack has an area that is twice

of A1, what will F2 be in comparison to F1
Physics
1 answer:
Firlakuza [10]2 years ago
7 0

Answer:

2F_{1}

Explanation:

F₁ = Force on one side of the jack

A₁ = Area of cross-section of one side of the jack

F₂ = Force on second side of the jack

A₂ = Area of cross-section of second side of the jack = 2 A₁

Using pascal's law

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}

F_{1}= \frac{F_{2}}{2}\\

F_{2}= 2F_{1}

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1. Determina el momento que produce una fuerza de 7 N tangente a una rueda de un metro de diámetro, sabiendo que el punto de apl
Rudik [331]

Answer:

τ= F r     into the blade

Explanation:

The moment of a force is defined by

         τ = F x r

where the bold indicates vectors

Let us write in the expression in magnitude

         τ = F r sin θ

in our case the force is tangent to the wheel therefore the angle between F and the radius is 90º, and the sin 90 = 1

       τ= F r

The direction of τ can be used by the rule of the right hand, the fingers curve in the direction of the torque when advancing from the force to the radius and the thumb points in the direction of the torque.

In this case, for a clockwise rotation, the fingers are curved in the direction and the thumb points into the blade, this is the direction of the τ.

TRASLATE

El momento de una fura es definido por

         τ = F x r

donde la negrillas indican vectores

Escribamos en ta expresión en magnitud

          τ = F r sin θ

en nuestro caso la fuerza es tangente a la rueda por lo tanto el angulo entre F y el radios es 90º, y el sin 90=1

        τ = F r

la dirección de tau la podemos  usar la regla de la mano derecha, los dedos curva en la dirección del torque al avanzar dese la fuerza al radio y el pulgar apunta en la dirección del torque.

En este caso para un giro en sentido horario los dedos se curvan ente sentido y el pulgar apunta hacia dentro de lla hoja, esta es la dirección del troque

5 0
2 years ago
A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy
GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

v=\sqrt{v_1^2+v_2^2}

v=\sqrt{5^2+8^2}=9.43\ m/s

Let E is the kinetic energy of the plane. It is given by :

E=\dfrac{1}{2}mv^2

E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

5 0
1 year ago
Read 2 more answers
What is the threshold frequency for sodium metal if a photon with frequency 6.66 × 1014 s−1 ejects a photon with 7.74 × 10−20 J
FrozenT [24]

Answer:

5.5 × 10^14 Hz or s^-1

no orange light has less frequency so no photoelectric effect

Explanation:

hf = hf0 + K.E

HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s

f is frequency of incident photon and f0 is threshold frequency

hf0 = hf- k.E

6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20

6.63 × 10 ^-34 × f0 = 3.64158×10^-19

                           f0 = 3.64158×10^-19/ 6.63 × 10 ^-34

                           f0 = 5.4925 × 10^14

                            f0 =5.5 × 10^14 Hz or s^-1

frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light

8 0
2 years ago
In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50μC and initially moving north
Goshia [24]

Answer:

a) The sign of the charge is positive.

b) The magnetic force on the particle is 0.050 newtons.

Explanation:

The magnetic force F on a moving charge with velocity v passing through a magnetic field B is:

\overrightarrow{F}=q\overrightarrow{v}\times\overrightarrow{B}(1)

a)

Because it is a cross product, we can find the direction of the force using the right-hand rule, that is too the direction of the movement. We have two possibilities here because the velocity vector and magnetic field are perpendicular: the particle deflects towards east or toward west, which depends on the charge of the particle. Note that if you put your right hand fingers, except thumb, pointing towards north (direction of velocity) and later close them in the direction of the magnetic field, if you maintain your thumb perpendicular to this movement it will point towards east (See figure), so that will be de direction of the force if the charge is positive, but if the charge is negative, the direction will be opposite (towards west). So the charge has to be positive to deflects towards east.

b)

Now by 1:

F=qvB\sin\theta=(8.50\times10^{-6})(4750)(1.25)\sin90\simeq\mathbf{0.050\,N}

5 0
2 years ago
To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch
sergij07 [2.7K]

Answer:

man upward acceleration is 0.14m/s^2

Explanation:

given data:

mass of man = 72 kg

downward force = 360 N

The mass of man of weight 72 kg is hang from two sections of rope, one section pf rope ties around man waist and other section is ties in man hands. when he pulls down the rope  with 360 N force then each section of  rope pulls with 360 N

we know that

Weight= mass × gravity= 72kg × 9.8 = 705.6N

Force = mass× acceleration

Force= -705.6 + (2 × 358) = 10.4 N

acceleration = \frac{10.4}{72} = 0.14m/s^2

4 0
2 years ago
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