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Amiraneli [1.4K]
3 years ago
5

A 1500 kg car rounds a horizontal curve with a radius of 52 m at a speed of 12 m/s. what minimum coefficient of friction must ex

ist between the road and tires to prevent the car from slipping?
Physics
1 answer:
Kitty [74]3 years ago
6 0

Answer:

The minimum coefficient of friction must exist between the road and tires to prevent the car from slipping is μ= 0.28

Explanation:

m= 1500 kg

r= 52m

Vt= 12m/s

g= 9.8 m/s²

Vt= ω * r

ω= Vt/r

ω= 0.23 rad/s

ac= ω²* r

ac= 2.77 m/s²

Fr= m* ac

Fr= 4153.84 N

W= m*g

W= 14700 N

Fr= μ * W

μ= Fr/W

μ=0.28

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Using monochromatic light of 410 nm wavelength, a single thin slit forms a diffraction pattern, with the first minimum at an ang
VLD [36.1K]

Answer:

c. 307 nm

Explanation:

angular position of first dark fringe =  λ / d ,   λ is wavelength and  d is width of slit .

(40 x π ) / 180 = 410 / d

angular position of second  dark fringe =  2 x λ / d ,   λ is wavelength and  d is width of slit .

(60 x π ) / 180 = 2 x λ / d

Dividing these equations

60 / 40 =  2 x λ / 410

λ = 307.5 nm.

5 0
3 years ago
FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine
Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

\tau=Fd=mg\cdot R

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

\tau=I\alpha

therefore, equating this to the above equation gives

mg\cdot R=I\alpha

solving for alpha gives

\alpha=\frac{mgR}{I}

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}\boxed{\alpha=1.47s^{-2}}

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

\theta=\frac{1}{2}\alpha t^2

Therefore, ar t = 4.2 s, the above gives

\theta=\frac{1}{2}(1.47)(4.2)^2

\boxed{\theta=12.97}

So how many revolutions is this?

To find out we just divide by 2 pi:

\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}\boxed{\#\text{rev}=2.06}

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

\omega^2=w^2_0+2\alpha(\Delta\theta)_{}

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

\omega^2=0+2(1.47)(12.97)w^2=38.12\boxed{\omega=6.17.}

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0
1 year ago
Match the organs to the functions they perform during digestion.<br> passes food to esophagus
mixas84 [53]

Answer:

the mouth I think

Because: the oesophagus is the throat and the mouth is how the food gets to the oesophagus. might be wrong because I'm not sure about the organ part but I'm pretty sure that's all it could be

8 0
3 years ago
Which form of the law of conservation of energy describes the motion of the block as it slides on the floor from the bottom of t
aivan3 [116]

Answer:

Explanation:

Let assume begins movement at zero point, that is, height is equal to zero. The block has an initial linear kinetic energy and no gravitational potential energy and end with no linear kinetic energy, some gravitational potential energy and work losses due to slide friction. In mathematical terms, this system can be model as follows:

K_{1} = U_{2} + W_{loss, 1 \longrightarrow 2}

Where K, U, W are linear kinetic energy, gravitational potential energy and work, respectively.

8 0
3 years ago
A force of 45 newtons is applied on an object, moving it 12 meters away in the same direction as the force. What is the magnitud
klio [65]

If the applied force is in the same direction as the object's displacement, the work done on the object is:

W = Fd

W = work, F = force, d = displacement

Given values:

F = 45N

d = 12m

Plug in and solve for W:

W = 45(12)

W = 540J

5 0
3 years ago
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