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1 year ago

What is the approximate increase in size from a 1 w to a 2w carbon resistor?

1 answer:

8 0

Since the 2-watt jobber has to dissipate twice as much power as the smaller unit I would expect it to have double the surface area. So if they are the usual cylindrical striped resistors then the larger one could have either double the length or else 1.4 times the radius of the smaller one.

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An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 2530 km. When the astronaut jumps upwa

Natali [406]

Answer:

1.38*10^18 kg

Explanation:

According to the Newton's law of universal gravitation:

$F=G*\frac{m_a*m_p}{r^2}$

where:

G= Gravitational constant (6.674×10−11 N · (m/kg)2)

ma= mass of the astronaut

mp= mass of the planet

$F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2$

so:

$m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg$

7 0

2 years ago

A body A of mass 1.5kg, travelling along the positive x-axis with speed 4.5m/s, collides with another body B of mass 3.2kg which

Pie

Momentum before the collision
x-direction:
p = m₁v₁ = 1.5 * 4.5 = 6.75
x-direction:
p = 0

momentum after the collision is conserved:
x-direction:
p = 6.75 = m₁v₁ + m₂v₂ = 1.5 * 2. 1* cos -30° + 3.2 * v₂*cos θ
y-direction:
p = 0 = m₁v₁ + m₂v₂ = 1.5 * 2.1 * sin -30° + 3.2 * v₂ * sin θ

Solve the two equations for v₂ and θ.

5 0

1 year ago

Suppose the foreman had released the box from rest at a height of 0.25 m above the ground. What would the crate's speed be when

Arturiano [62]

Answer:

v = 2.21 m/s

Explanation:

The foreman had released the box from rest at a height of 0.25 m above the ground.

We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.25} \\\\v=2.21\ m/s$

So, its velocity at the bottom of the ramp is 2.21 m/s.

4 0

1 year ago

Calculate the flux of the vector field F⃗ =−6i⃗ +5x2j⃗ −5k⃗ , through the square of side 8 in the plane y=1, centered on the y-a

Tasya [4]

Answer:

The flux is 682.6 Wb.

Explanation:

Given that,

Vector field $F=-6i+5x^2j-5k$

We need to calculate the flux

Using formula of flux

$\phi=\int_{-4}^{4}\int_{-4}^{4}(F\cdot j\ dxdz)$

Put the value into the formula

$\phi=\int_{-4}^{4}\int_{-4}^{4}(-6i+5x^2j-5k)1\ dxdz$

$\phi=\int_{-4}^{4}\int_{-4}^{4}(5x^2)dxdz$

$\phi=2(\dfrac{x^3}{3})_{-4}^{4}\times(z)_{-4}^{4}$

$\phi=682.6\ Wb$

Hence, The flux is 682.6 Wb.

7 0

1 year ago

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