1.7 x 10^2 N
or 166 N
First you find the vertical component of the weight, which is 9.8*40, (g*m), which is 392 N. You then find the angle between that and the slope, which is 90-25, which is 65. You then multiply the vertical weight by cos(65), to find the component of that that is parallel to the slope. You get 165.666 N
Answer:
313.92w
Explanation:
Formula for power:
P=W/∆t = Fv
Givens:
m=20kg
∆y=4.0m
∆t=2.5s
a=9.81m/s²
In order to find power, we first need to solve for work.
W=Fd (force*displacement), f=mg
W=mg∆y
W=(20kg)(9.81m/s²)(4.0m)
W=784.8J
P=W/∆t
P=784.8J/2.5s
P=313.92 watts
Answer:
If one end of a metal bar is heated, the atoms at that end vibrate more than the atoms at the cold end. The vibration spreads along the bar from atom to atom.
Explanation:
The spread of heat in this way is called conduction. Metals are good conductors of heat.
Answer:
Its heat capacity is higher than that of any other liquid or solid, its specific heat being 1 cal / g, this means that to raise the temperature of 1 g of water by 1 ° C it is necessary to provide an amount of heat equal to a calorie . Therefore, the heat capacity of 1 g of water is equal to 1 cal / K.
Explanation:
The water has a very high heat capacity, a large amount of heat is necessary to raise its temperature 1.0 ° K. For biological systems this is very important because the cellular temperature is modified very little in response to metabolism. In the same way, aquatic organisms, if water did not possess that quality, would be very affected or would not exist.
This means that a body of water can absorb or release large amounts of heat, with little temperature change, which has a great influence on the weather (large bodies of water in the oceans take longer to heat and cool than the ground land). Its latent heats of vaporization and fusion (540 and 80 cal / g, respectively) are also exceptionally high.
Answer:

Explanation:
v = Velocidad final = 
u = Velocidad inicial = 0
t = Tiempo empleado = 15 s
a = Aceleración
De las ecuaciones cinemáticas tenemos

La aceleración del camión en el primer intervalo de tiempo es
.