I believe it would best represent Newton’s first law; an object tends to stay at rest and an object tends to stay in motion unless acted upon by an unbalanced force. When the dog stops walking, the doll will continue to go forward because there is no unbalanced force acting in it.
Answer:
2.61 atm
Ley de Boyle
Explanation:
= Presión inicial = 0.96 atm
= Presión final
= Volumen inicial = 95 mL
= Volumen final = 35 mL
En este problema usaremos la ley de Boyle.
La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.
You need to figure out t4 to know the tension in the string.
Since the whole thing is not moving t1 + t2 + t3 = t4.
torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)
t1 =3.2 * 44g
t2 = 7 * 49g
t3 = 3.5 * 24g
t4 = t1 + t2 + t3 = 5570,118
The t4 also is given by:
t4 = r * T * sin Ф
r = 7
Ф = 32°
T: tension in the string
T = t4 / (r * sinФ)
T = t4 / (7 * sin(32°))
T = 1501,6 N
Answer:
2.10L
Explanation:
Given data
V1= 2.5L
T1= 275K
P1= 2.1atm
P2= 2.7 atm
T2= 298K
V2= ???
Let us apply the gas equation
P1V1/T1= P2V2/T2
substitute into the expression we have
2.1*2.5/275= 2.7*V2/298
5.25/275= 2.7*V2/298
Cross multiply
275*2.7V2= 298*5.25
742.5V2= 1564.5
V2= 1564.5/742.5
V2= 2.10L
Hence the final volume is 2.10L
