Ask question

Ask question

All categories

3 years ago

8

What is the maximum speed with which a 1200-kg car can round a turn of radius 90.0 m on a flat road if the coefficient of fricti

on between tires and road is 0.65? Is this result independent of the mass of the car?

1 answer:

NemiM [27]3 years ago

4 0

No it doesn't depend on the mass of the car
the maximum speed will be = √urg
√0.65*90*9.8 = v
which comes around 24 m/s

You might be interested in

The impulse given to a ball with mass of 2 kg is 16 N*s. If the ball starts from rest, what is its final velocity?

Alik [6]

The impulse is equal to the variation of momentum of the object:
$I=\Delta p = m \Delta v$
where m is the mass object and $\Delta v = v_f - v_i$ is the variation of velocity of the object.

The ball starts from rest so its initial velocity is zero: $v_i=0$ . So we can rewrite the formula as
$I=m v_f$
or
$v_f = \frac{I}{m}$

and since we know the impulse given to the ball (I=16 Ns) and its mass (m=2 kg), we can find the final velocity of the ball:
$v_f = \frac{16 Ns}{2 kg}= 8 m/s$

7 0

3 years ago

Read 2 more answers

A small button placed on a horizontal rotating platform with diameter 0.520 m will revolve with the platform when it is brought

yawa3891 [41]

For the position of button we can use force balance as

Friction Force = Centripetal force

so here we will have

$\mu_s mg = m\omega^2 R$

here we know that

R = radius of circle where button is placed

$\omega = 2\pi f$

f = 35 rev/min

$f = \frac{35}{60} rev/s = 0.58 rev/s$

$\omega = 2\pi (0.58) = 3.66 rad/s$

now from above equation

$\mu_s(m)(9.81) = (3.66)^2(0.240)$

$\mu_s = 0.33$

so friction coefficient will be 0.33

5 0

4 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

3 years ago

A car starting from rest accelerates in a straight line at a constant rate of 5.5m/s for 6s.If the car after this acceleration s

aalyn [17]

Answer:

The time it takes to stop is 13.75 seconds

Explanation:

A body moving with constant acceleration, 'a', for a time, 't', has a final velocity, 'v', given by the following kinematic equation;

v = u + a·t

Where;

v = The final velocity of the body

a = The acceleration of the body

t = The time of acceleration (accelerating period) of the body

u = The initial velocity of the body

The given parameters for the acceleration of the car are;

The initial velocity of the car, u = 0 m/s (a car starting from rest)

The constant acceleration of the car, a = 5.5 m/s²

The acceleration duration, t = 6 s

Therefore, we have;

The final velocity of the car after the acceleration, v = 0 m/s + 5.5 m/s² × 6 s = 33 m/s

The final velocity of the car after the acceleration, v = 33 m/s

When the car slows down uniformly, and comes to a stop (final velocity, v₂ = 0 m/s), it has a constant negative acceleration, (deceleration) '-a₂'

The given parameters when the car slows down are;

The deceleration, -a₂ = 2.4 m/s²

The final velocity, v₂ = 0 m/s

The initial velocity, u₂ = v = 33 m/s

The time it takes to stop = t₂

-a₂ = 2.4 m/s²

∴ a₂ = -2.4 m/s²

From, v = u + a·t, we have;

v₂ = v + a₂·t₂

By plugging in the values of the variables, we have;

0 m/s = 33 m/s + (-2.4 m/s²) × t₂

∴ 2.4 m/s² × t₂ = 33 m/s

t₂ = 33 m/s/(2.4 m/s²) = 13.75 s

The time it takes to stop, t₂ = 13.75 seconds

7 0

3 years ago

An 8 g bullet leaves the muzzle of a rifle with

Elena-2011 [213]

Answer: 1872 N

Explanation:

This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:

$V^{2}=V_{o}^{2} + 2ad$ (1)

$F=ma$ (2)

Where:

$V=611.9 m/s$ is the bullet's final speed (when it leaves the muzzle)

$V_{o}=0$ is the bullet's initial speed (at rest)

$a$ is the bullet's acceleration

$d=0.8 m$ is the distance traveled by the bullet before leaving the muzzle

$F$ is the force

$m=8 g \frac{1 kg}{1000 g}=0.008 kg$ is the mass of the bullet

Knowing this, let's begin by isolating $a$ from (1):

$a=\frac{V^{2}}{2d}$ (3)

$a=\frac{(611.9 m/s)^{2}}{2(0.8 m)}$ (4)

$a=234013.5063 m/s^{2} \approx 2.34(10)^{5} m/s^{2}$ (5)

Substituting (5) in (2):

$F=(0.008 kg)(2.34(10)^{5} m/s^{2})$ (6)

Finally:

$F=1872 N$

4 0

3 years ago