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3 years ago

si un disco fue lanzado con una fuerza de 1000 N, y el disc recorrió una distancia e o.6 m ¿Qué trabajo efectuó el lanzador ?

Physics

1 answer:

Alexeev081 [22]3 years ago

4 0

Answer:

Trabajo realizado = 600 Nm

Explanation:

Dados los siguientes datos;

Fuerza = 1000 Newton

Distancia = 0.6 metros

Para encontrar el trabajo realizado;

$Trabajo \; realizado = fuerza * distancia$

Sustituyendo en la ecuación, tenemos;

$Trabajo \; realizado = 1000 * 0.6$

Trabajo realizado = 600 Nm

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A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surf

Sati [7]

Answer:

a) The minimum thickness of the oil slick at the spot is 313 nm

b) the minimum thickness be now will be 125 nm

Explanation:

Given the data in the question;

a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

t $_{min$ = λ/2n

given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20

we substitute

t $_{min$ = 750 / 2(1.20)

t $_{min$ = 750 / 2.4

t $_{min$ = 312.5 ≈ 313 nm

Therefore, The minimum thickness of the oil slick at the spot is 313 nm

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

minimum thickness of the oil slick at the spot will be;

t $_{min$ = λ/4n

given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50

we substitute

t $_{min$ = 750 / 4(1.50)

t $_{min$ = 750 / 6

t $_{min$ = 125 nm

Therefore, the minimum thickness be now will be 125 nm

4 0

3 years ago

What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem

8090 [49]

Answer:

$\Delta L = 4.2 m$

Explanation:

As per the formula of thermal expansion we know that

$L = L_o(1 + \alpha\Delta T)$

so here we will have

$L_o = 1400 m$

$\alpha = 12 \times 10^{-6} per ^oC$

$\Delta T = 250 degree C$

so here change in the length of the rod is given as

$\Delta L = L - L_o$

$\Delta L = L_o \alpha \Delta T$

$\Delta L = 1400 (12 \times 10^{-6})(250)$

$\Delta L = 4.2 m$

8 0

4 years ago

A projectile is fired with an initial velocity of 120.0 meters per second at an angle, θ above the horizontal. If the projectile

k0ka [10]

Answer:

θ = 62.72°

Explanation:

The projectile describes a parabolic path:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = x₀+ vx*t Formula (1)

vx = v₀x

Where:

x: horizontal position in meters (m)

x₀: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m/s

v₀x: Initial speed in x in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:

y= y₀+(v₀y)*t - (1/2)*g*t² Formula (2)

vfy= v₀y -gt Formula (3)

Where:

y: vertical position in meters (m)

y₀ : initial vertical position in meters (m)

t : time in seconds (s)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

Data

v₀ = 120 m/s , at an angle θ above the horizontal

v₀x= 55 m/s

x-y components of the initial velocity ( v₀)

v₀x = v₀*cosθ Equation (1)

v₀y = v₀*sinθ Equation (2)

Calculating of the angle θ

We replace data in the Equation (1)

55 = 120*cosθ

cosθ = 55 / 120

$\theta = cos^{-1}( \frac{55}{120} )$

θ = 62.72°

3 0

3 years ago

Inside a television picture tube there is a build-up of electrons (charge of 1.602 × 10^–19 C) with an average spacing of 38.0 n

Brut [27]

The magnitude of electric field is produced by the electrons at a certain distance.

E = kQ/r²

where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9 N. m2 / C2

Given are the following:
Q = 1.602 × 10^–19 C
r = 38 x 10^-9 m

Substitue the given:
E = $\frac{( 1.602 x 10^{-19} )( 9.0x10_{9} )}{(38x10^{-9}) ^{2} }$

E = 998.476 kN/C

8 0

3 years ago

Read 2 more answers

In doing a load of clothes, a clothes dryer uses 16 A of current at 240 V for 45 min. A personal computer, in contrast, uses 2.7

EastWind [94]

Answer:

(a) Total time will be 8.05 hour

(B) Total cost will be $417.6

Explanation:

We have given dryer uses 16 A of current so current i = 16 A

Voltage V = 240 volt

Time t = 45 minutes =45×60 = 2700 sec

So power P = VI = 240×16 = 3840 W

So energy E = power×time = 3840×45×60 = 9396 KJ

(A) Now current in the computer i = 2.7 A

Voltage V = 120 V

So power P = 120×2.7 = 324 W

Now computer is using energy of dryer

So time by which it will use energy is $t=\frac{9396000}{324}=29000sec=8.05hour$

So time will be 8.05 hour

(b) Cost per kilowatt is $0.12

So total cost =3480×0.12 = $417.6

4 0

3 years ago