Raul empuja su auto de 950 KG sobre una pista rugosa e inclinada 15,0 con una fuerza paralela a la pista, cuya magnitud es de 200 N. Determina el trabajo mecanico hecho por el peso sobre el auto para un desplazamiento de 5,00m (gravedad=10m/5)

966 joules

Step-by-step explanation:

Step one:

given data

mass of car= 950kg

angle of incline= 15°

Applied force= 200N

displacement = 5m

If we apply a force on an object through a distance work is done, hence Mechanical work is the product of a force applied on an object and its displacement while that force is applied

Step two:

Work done= Force * distance

Since the car is moved on an inclined plane, the expression for the work done is

Work-done= Frcos∅

where r is the distance moved

Work-done= 200*5cos 15

Work-done= 200*4.83

Work-done= 966 joules

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2