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galina1969 [7]
2 years ago
12

Un pintor de 75.0 kg sube por una escalera de 2.75 m que está inclinada contra una pared vertical. La escalera forma un ángulo d

e 30.0° con la pared. ¿Cuánto trabajo (en Julios) realiza la gravedad sobre el pintor? *
Physics
1 answer:
dezoksy [38]2 years ago
6 0

Answer:

Work done, W = 1786.17J

Explanation:

The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "

Mass of a painter, m = 75 kg

He climbs 2.75-m ladder that is leaning against a vertical wall.

The ladder makes an angle of 30 degrees with the wall.

We need to find the work done by the gravity on the painter.

The angle between the weight of the painter and the displacement is :

θ = 180 - 30

= 150°

The work done by the gravity is given by :

W=Fd\cos\theta\\\\=75\times 10\times 2.75\times \cos30\\\\=1786.17\ J

Hence, the required work done is 1786.17 J.

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distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm.
a_sh-v [17]

Answer:

The speed of the galaxy relative to the Earth is 3.09\times 10^6\ m/s.

Explanation:

We have,

(a) Wavelength emitted by light at distant galaxy is 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. It can be seen that the wavelength of light reduces as it reaches Earth. It is called Red shift. As per Doppler's effect, we can say that the galaxy is receding from the Earth.

(b) Let v is the speed of the galaxy relative to the Earth. It can be given by :

v=c(\dfrac{\lambda'}{\lambda}-1)\\\\v=3\times 10^8\times (\dfrac{438.6 }{434.1 }-1)\\\\v=3\times 10^8\times (\dfrac{438.6}{434.1}-1)\\\\v=0.0103\cdot3\cdot10^{8}\\\\v=3.09\times 10^6\ m/s

So, the speed of the galaxy relative to the Earth is 3.09\times 10^6\ m/s.

4 0
3 years ago
A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse
Elanso [62]

Answer:

v_{f} = 0.51 \frac{m}{s}

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg :  crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight  (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)   

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W = 0

N = W

N = 882 N

We replace the  data in the formula (2)

Ff= μk*N  = 0.351* 882 N

Ff=  309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax    , ax=0

282 N - 309.58 N = 90*a  

a=  (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the  data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}

v_{f} = 0.51 \frac{m}{s}

8 0
3 years ago
A current exists whenever electric charges move. If ΔQ is the net charge that passes through a surface during a time period Δt,
jeka57 [31]

Answer:

It represents the change in charge Q from time t = a to t = b

Explanation:

As given in the question the current is defined as the derivative of charge.

                                  I(t) = dQ(t)/dt ..... (i)

But if we take the inegral of the equation (i) for the time interval  from t=a to

t =b we get

                                   Q =∫_a^b▒〖I(t)  〗 dt

which shows the change in charge Q from time t = a to t = b. Form here we can say that, change in charge is defiend as the integral of current for specific interval of time.

5 0
3 years ago
4. Compara. La composición de la gasolina para los coches cambia del invierno al verano. La mezcla de los componentes de la gaso
dedylja [7]
Skisosjssnbxhxsndjsksksksa
5 0
3 years ago
Am i correct? If not then which one
cupoosta [38]

Answer:

Yes, it's correct

Explanation:

Newton's second Law states that the acceleration of an object is proportional to the net force applied on it, according to the equation:

F=ma

where

F is the net force on the object

m is the mass of the object

a is the acceleration of the object

We can re-arrange the previous equation in order to solve explicitely for a, the acceleration, and we find:

F=ma\\\frac{F}{m}=\frac{ma}{m}\\\frac{F}{m}=a\\a=\frac{F}{m}

So, we see that the acceleration is proportional to the net force and inversely proportional to the mass of the object.

4 0
3 years ago
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