The power in an electrical circuit is given by the equation P= RR, where /is

A particle moving along the x-axis has a position given by x = (24t – 2.0t 3 ) m, where t is measured in s. What is the magnitud

Vitek1552 [10]

<h2>The magnitude 24 ( $\dfrac{m}{s^2}$ ) of the acceleration of the particle when the particle is not moving.</h2>

Explanation:

Given,

A particle moving along the x-axis has a position given by

$x=(24t-2.0t^3)$ m ........ (1)

To find, the magnitude ( $\dfrac{m}{s^2}$ ) of the acceleration of the particle when the particle is not moving = ?

Differentiating equation (1) w.r.t, 't', we get

$\dfrac{dx}{dt} =\dfrac{d((24t-2.0t^3))}{dt}$

⇒ $\dfrac{dx}{dt} =24(1)-3(2.0)t^{2} =24-6t^{2}$ ....... (2)

⇒ $24-6t^{2} = 0$

⇒ $t^{2}=2^{2}$

⇒ t = 2 s

Again, differentiating equation (2) w.r.t, 't', we get

$\dfrac{d^2x}{dt^2} =-12t$

Put t = 2, we get

$\dfrac{d^2x}{dt^2} =-12(2)=24$

Thus, the magnitude 24 ( $\dfrac{m}{s^2}$ ) of the acceleration of the particle when the particle is not moving.

3 0

3 years ago

Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem

statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

$x=x_o+v_1t$ (1)

xo = 10 km

v1 = 55km/h

car 2:

$x'=v_2t$ (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

$x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}$

$t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s$

The position in which both cars coincides is:

$x=(55km/h)(0.33h)=18.33km$

6 0

2 years ago

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

m_a_m_a [10]

Answer:0.27

Explanation:

Given

One worker Pushes with force $F_1=430 N$

other Pulls it with a rope of rope $F_2=360 N$

mass of crate $m=290 kg$

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

$f_r=F_1+F_2$

where $f_r$ is the friction force

$f_r=430+360$

$f_r=790 N$

$f_r=\mu N$

where $\mu$ is the coefficient of static friction

$N=mg$

$790=\mu 29\cdot 9.8$

$\mu =0.27$

6 0

3 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is