Answer:
1,85 m / s²
Explanation:
De la pregunta anterior, se obtuvieron los siguientes datos:
Velocidad inicial (u) = 40 km / h
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:
1 km / h = 0,2778 m / s
Por lo tanto,
40 km / h = 40 km / h × 0,2778 m / s / 1 km / h
40 km / h = 11,11 m / s
Por tanto, 40 km / h equivalen a 11,11 m / s.
Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:
Velocidad inicial (u) = 11,11 m / s
Hora inicial (t₁) = 0
Tiempo final (t₂) = 6 s
Velocidad final (v) = 0
Aceleración (a) =?
a = (v - u) / (t₂ - t₁)
a = (0 - 11,11) / (6 - 0)
a = - 11,11 / 6
a = –1,85 m / s²
Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²
Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
Hi, thank you for posting your question here at Brainly.
Newton's second law of motion can be expressed as Fnet = ma. The next external for acting on, say for example, a moving car are the following:
*weight due to gravity (force down)
*friction force between he road and the car's tires (force opposite the car's direction)
Answer:
Explanation:
Question 1
An arrow weighing 20g shortly after firing has a speed of 50m / s. Calculate the work done by the athlete. What is the potential energy of the elasticity of the tensed string?
mass m = 20g = 20/1000 = 0.02kg
speed v = 50m / s
P.E = K.E = ½mv²
P.E = ½ × 0.02 × 50²
P.E = 25 J
work done = P.E = 25J
Qestion 2
A 80 kg athlete stood on a trampoline with a coefficient of elasticity of k = 2 kN / m. As far as the edge of the trampoline lowers.
force of elasticity
F = -kx
x = F / k
in our case F will be the force of pressure or gravity
F = mg
g is gravitational acceleration, and according to Newton's second law, acceleration is force through mass - unit of force N, unit of mass kg. Acceleration either in m / s ^ 2 or N / kg
F = 80kg * 10N / kg = 800 N
x = 800N / -2000N = -0.4
The trampoline will lower, so from the level by 0.4 meters and hence this minus
