The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
Responder:
6.704 m / s
Explicación:
Se dice que el trabajo se realiza cuando la fuerza aplicada a un objeto hace que el objeto se mueva. Primero necesitamos calcular la distancia recorrida por el perro usando la fórmula del trabajo realizado.
Trabajo realizado = Fuerza × distancia
Distancia = Trabajo realizado / Fuerza
Distancia = W / mg
S = 176/8 × 9,81
S = 176 / 78,48
S = 2,24 m
Dada la velocidad inicial u = 3.6km / h
Convertir a m / s
= 3.6km × 1000m / 1h × 3600
= 3600/3600
= 1 m / s
u = 1 m / s
Usando la ecuación de movimiento
v² = u² + 2gS para obtener la velocidad final v:
v² = 1² + 2 (9,81) (2,24)
v² = 1 + 43,9488
v² = 44,9488
v = √44,9488
v = 6,704 m / s
Por tanto, la rapidez final del perro es de 6,704 m / s
Answer : The final volume of the balloon at this temperature and pressure is, 17582.4 L
Solution :
Using combined gas equation is,
where,
= initial pressure of gas = 1 atm
= final pressure of gas = 0.3 atm
= initial volume of gas = 6000 L
= final volume of gas = ?
= initial temperature of gas = 273 K
= final temperature of gas = 240 K
Now put all the given values in the above equation, we get the final pressure of gas.
Therefore, the final volume of the balloon at this temperature and pressure is, 17582.4 L
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