What is the angular acceleration experienced by a uniform solid disc of mass 5-kg and radius 20 cm when a net

B. A car moving at an initial speed vi applies its brakes and skids for some distance until coming to a complete stop. If the co

wariber [46]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The distance which the car skid is $l = \frac{v_i^2 }{2 * \mu_k * g }$

Explanation:

From the question we are told that

The initial velocity of the car is $v_i$

The coefficient of kinetic friction is $\mu_k$

According to the law of energy conservation

The initial Mechanical Energy = The final Mechanical Energy

$M_i = M_f$

The initial mechanical energy is mathematically represented as

$M_i = KE _o + PE_e$

where KE is the initial kinetic energy which is mathematically represented as

$KE = \frac{1}{2} m v_i^2$

And PE is the initial potential energy which is zero given that the car is on the ground

Now

$M_f = W_{\mu}$

Where $W_{\mu}$ is the work which friction exerted on the car which is mathematically represented as

$W_{\mu} = m* \mu_k * g * l$

Where $l$ is the distance covered by the car before it slowed down

$\frac{1}{2} m v_i^2 = m* \mu_k * g * l$

=> $l = \frac{v_i^2 }{2 * \mu_k * g }$

3 0

4 years ago

A cheetah runs at 93km/he to the east. This describes the cheetah’s

Ainat [17]

The answer is average speed

8 0

3 years ago

At a certain instant, a rotating turbine wheel of radius RR has angular speed ωω (measured in rad/srad/s). What must be the magn

jonny [76]

Answer:

Explanation:

The magnitude of the acceleration makes an angle of 30° with the tangential velocity.

Resolving the acceleration to tangential and radial acceleration

at = aCos30 = √3a/2

ar = aSin30 = ½a

a = 2•ar

Then, the tangential acceleration is the linear acceleration, so the relationship between the tangential acceleration and angular acceleration is given as:

at = Rα

Then, α = at/R

since at = √3a/2

Then, α = √3 at/2R, equation 1

The radial acceleration is given as

ar = ω²R

Note that, at² + ar² = a²

at = √(a²-ar²)

Back to equation 1

α = √3 at/2R

α = √3√(a²-ar²)/2R

α = √3√(a²-(w²R)²)/2R

α = √3(a²-w⁴R²) / 2R

Also, a = 2•ar = 2w²R

Then,

α = √3((2w²R)²-w⁴R²) / 2R

α = √3(4w⁴R²-w⁴R²) / 2R

α = √3(3w⁴R²) / 2R

α = √9w⁴R² / 2R

α = 3w²R / 2R

α = 3w²/2

7 0

3 years ago

The following represents a mass attached to a spring oscillating in simple harmonic motion. X(t) = 4.0 cos(3.0t +0.10) units of

kolbaska11 [484]

Answer:

a) A = 4.0 m , b) w = 3.0 rad / s , c) f = 0.477 Hz , d) T = 20.94 s

Explanation:

The equation that describes the oscillatory motion is

x = A cos (wt + fi)

In the exercise we are told that the expression is

x = 4.0 cos (3.0 t + 0.10)

let's answer the different questions

a) the amplitude is

A = 4.0 m

b) the frequency or angular velocity

w = 3.0 rad / s

c) angular velocity and frequency are related

w = 2π f

f = w / 2π

f = 3 / 2π

f = 0.477 Hz

d) the period

frequency and period are related

T = 1 / f

T = 1 / 0.477

T = 20.94 s

e) the phase constant

Ф = 0.10 rad

f) velocity is defined by

v = dx / dt

v = - A w sin (wt + Ф)

speed is maximum when sine is + -1

v = A w

v = 4 3

v = 12 m / s

g) the angular velocity is

w² = k / m

k = m w²

k = 1.2 3²

k = 10.8 N / m

h) the total energy of the oscillator is

Em = ½ k A²

Em = ½ 10.8 4²

Em = 43.2 J

i) the potential energy is

Ke = ½ k x²

for t = 0 x = 4 cos (0 + 0.1)

x = 3.98 m

j) kinetic energy

K = ½ m v²

for t = 00.1 ²

v = A w sin 0.10

v = 4 3 sin 0.10

v = 1.98 m / s

3 0

3 years ago

Will mark brainliest if answer

Anvisha [2.4K]

I won't touch the object because there is a warning sign and that means you stay out if it because it is a danger sign that High Resistance Material that means it is dangerous.

3 0

2 years ago