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2 years ago

A car of mass 1500 kg is negotiating a flat circular curve of radius 50 m with a speed of 20 m/s.

Physics

1 answer:

Lilit [14]2 years ago

6 0

Answer:

Explanation:

a. The source of centripetal force on the car is (3) the static friction force.

b. ac = v²/R = (20²)/50 = 8 m/s²

c. Fc = m(ac) = 1500(8) = 12 kN

d. μ = Fc/N = Fc/mg = 12000 / 1500(9.8) = 0.8163... ≈ 0.82

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Please help what is the definition ligthspeed

aev [14]

Light speed is the speed at which light can travel in a "vacuum" (space for example).

The speed of light is constant and is exactly: 299,792,458 meters per second

ONLY IN A VACUUM!!

The speed of light changes when it goes through different mediums (such as from space to Earth, light travels slower in Earth).

Good luck!

3 0

2 years ago

A 12.0N force with a fixed orientation does work on a

kvasek [131]

Answer:

(a) $\theta=62.31^{\circ}$

(b) $\theta=117.68^{\circ}$

Explanation:

It is given that,

Force acting on the particle, F = 12 N

Displacement of the particle, $d=(2.00i -4.00j+3.00k)\ m$

Magnitude of displacement, $d=\sqrt{2^2+4^2+3^2}= 5.38\ m$

(a) If the change in the kinetic energy of the particle is +30 J. The work done by the particle is given by :

$W=Fd\ cos\theta$

$\theta$ is the angle between force and the displacement

According to work energy theorem, the charge in kinetic energy of the particle is equal to the work done.

So,

$cos\theta=\dfrac{W}{Fd}$

$cos\theta=\dfrac{+30\ J}{12\times 5.38}$

$\theta=62.31^{\circ}$

(b) If the change in the kinetic energy of the particle is (-30) J. The work done by the particle is given by :

$cos\theta=\dfrac{W}{Fd}$

$cos\theta=\dfrac{-30\ J}{12\times 5.38}$

$\theta=117.68^{\circ}$

Hence, this is the required solution.

8 0

2 years ago

What two elements are named after Dimitri Mendeleev?

bezimeni [28]

I'm pretty sure there is only one element named after Mendeleev: <span>Mendelevium.</span>

8 0

2 years ago

Read 2 more answers

f the CD rotates clockwise at 500 rpmrpm (revolutions per minute) while the last song is playing, and then spins down to zero an

vazorg [7]

Answer:

The magnitude of the angular acceleration is $a = 20.14 rad/s^2$

Explanation:

From the question we are told that

The angular speed of CD is $w_{CD} = 500 rpm = \frac{500 rpm}{\frac{60 \ s }{1 \ min} } * \frac{2 \pi }{ 1 \ rev} = 52.37 rad/s$

time taken to decelerate is $t_{CD} = 2.60\ s$

The final angular speed is $w_f= 0 \ rad/s$

The angular acceleration is mathematically represented as

$a = \frac{w_f - w_{CD}}{t}$

substituting values

$a = \frac{0 - 52.37}{2.60}$

$a = - 20.14 rad/s^2$

The negative sign show that the CD is decelerating but the magnitude is

$a = 20.14 rad/s^2$

3 0

2 years ago

A 4.00-kg particle moves along the x axis. Its position varies with time according to x 5 t 1 2.0t 3, where x is in meters and t

White raven [17]

Answer:

Explanation:

Given equation is ,

x =t + 2 t³ ,

dx/dt = velocity ( v ) = 1 + 6 t²

a) kinetic energy = 1/2 m v² = .5 x 4 ( 1 + 6 t² )² = 2 ( 1 + 6 t²)²

b ) Acceleration = dv /dt = 12 t .

force( F ) = mass x acceleration = 4 x 12 t = 48 t

Power = force x velocity = 48 t x ( 1 + 6 t²). = 48 t + 288 t³ )

work done = ∫ F dx =∫ 48 t x( 1 + 6t² )dt ; = [48t²/2 + 48 x 6 x t³ /3 = 24 t² + 96 t³ )]₀² = 864 J

6 0

2 years ago

Read 2 more answers