Answer:
ans:
tenson(T) = 20 N
acceleration (a) = 2.86 m/s
Explanation:
T + mg = Mg
T = Mg - mg
T = g( M - m )
T = 10× ( 7-5 )
T = 20 N
again;
T = 20
Ma = 20
a = 20 / 7
= 2.86 m/s
¿Es esta una pregunta verdadera o falsa? Por cierto, no hay mucha gente que hable español en esta aplicación, así que buena suerte.
Answer:
32 °C.
Explanation:
Hola.
En este caso, debemos entender que la relación entre el calor y la temperatura viene dada por:
De este modo, dado que estamos estudiando la misma sustancia (agua) con masa constante, la relación calor-temperatura es lineal y directamente proporcional, por tal razón, si se duplica el calor suministrado, la temperatura también será duplicada, de modo que:
¡Saludos!
Answer:
0.49m
Explanation:
So you need to change the original equation for finding fields to find distance, and then just plug in the numbers
Which equals 0.49meters
Also it was right on Acellus :)
Hope this helps
Answer:
Explanation:
Sam mass=75kg
Height is 50m
20° frictionless slope
Horizontal force on Sam is 200N
According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.
Therefore
Wg - Ww =∆K.E
Note initial the body was at rest at top of the slope.
Then, ∆K.E is K.E(final) - K.E(initial)
K.E Is given as ½mv²
Since initial velocity is zero then, K.E(initial ) is zero
Therefore, ∆K.E=½mVf²
Wg is work done by gravity and it is given by using P.E formulas
Wg=mgh
Wg=75×9.8×50
Wg=36750J
Ww is work done by wind and it's is given by using formulae for work
Work=force × distance
Ww=horizontal force × horizontal distance
Using Trig.
TanX=opposite/adjacent
Tan20=h/x
x=h/tan20
x=50/tan20
x=137.37m
Then,
Ww=F×x
Ww=200×137.37
We=27474J
Now applying the formula
Wg - Ww =∆K.E
36750 - 27474 =½×75×Vf²
9276=37.5Vf²
Vf²=9275/37.5
Vf²= 247.36
Vf=√247.36
Vf=15.73m/s
