Ask question

Ask question

All categories

1 month ago

6

A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2. During a 4.20-s time inte

rval, the wheel rotates through 62.4 rad. What is the angular speed of the wheel at the end of the 4.20-s interval?

1 answer:

Yuliya22 [3.3K]1 month ago

4 0

We will use the equations of rotational kinematics,

$\theta =\theta _{0} + \omega_{0} t+ \frac{1}{2}\alpha t^2$ (A)

$\omega^2= \omega^2_{0} +2\alpha\theta$ (B)

Here, $\theta$ and $\theta _{0}$ denote the final and initial angular displacements, respectively, whereas $\omega$ and $\omega_{0}$ represent final and initial angular velocities, and $\alpha$ is the angular acceleration.

We are provided with $\alpha = - 25.60 \ rad/s^2, \theta = 62.4 \ rad$ and $t = 4.20 \ s$ .

By substituting these values into equation (A), we have

$62.4 \ rad = 0 + \omega_{0} 4.20 \ s + \frac{1}{2} (- 25.60 \ rad/s^2) ( 4.20)^2 \\\\ \omega_{0} = \frac{220.5+ 62.4 }{4.20} =67.4 \ rad/s$

Now, using equation (B),

$\omega^2=(67.4 \ rad/s)^2 + 2 (- 25.60 \ rad/s^2)62.4 \ rad \\\\\ \omega = 36.7 \ rad/s$

This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.

You might be interested in

If a 50.0-kg mass weighs 554 n on the planet saturn, calculate saturn’s radius

ValentinkaMS [3465]

Answer:

17.35 × 10^(-6) m

Explanation:

Mass; m = 50 kg

Weight; W = 554 N

From the formula:

W = mg

This simplifies to; 554 = 50g

g = 554/50

g = 11.08 m/s²

Also, using the formula;

mg = GMm/r²

hence; g = GM/r²

Rearranging gives;

r = √(GM/g)

With G as a known constant of 6.67 × 10^(-11) Nm²/kg²

r = √(6.67 × 10^(-11) × 50/11.08)

r = 17.35 × 10^(-6) m

8 0

1 month ago

A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

serg [3582]

Answer:

$v = 3369.2 m/s$

Explanation:

The beacon is rotating at an angular speed of

$f = 10 rev/min$

so we have

$\omega = 2\pi f$

$\omega = 2\pi(\frac{10}{60})$

$\omega = 1.047 rad/s$

We know that

$v = r \omega$

At this point we have

$r = 2 miles = 2(1609 m)$

$r = 3218 m$

So we can conclude with

$v = 3218(1.047)$

$v = 3369.2 m/s$

6 0

1 month ago

A cylinder has 500 cm3 of water. After a mass of 100 grams of sand is poured into the cylinder and all air bubbles are removed b

Keith_Richards [3271]

Answer: SG = 2.67

The specific gravity for the sand is 2.67

Explanation:

Specific gravity is determined by the formula: density of the substance/density of water

Provided information;

Mass of sand m = 100g

The volume of sand equals the volume of water it displaces

Vs = 537.5cm^3 - 500 cm^3

Vs = 37.5cm^3

Calculating density of sand = m/Vs = 100g/37.5 cm^3

Ds = 2.67g/cm^3

Density of water Dw = 1.00 g/cm^3

Thus, the specific gravity of the sand can be expressed as

SG = Ds/Dw

SG = (2.67g/cm^3)/(1.00g/cm^3)

SG = 2.67

The specific gravity of the sand stands at 2.67

3 0

17 days ago

A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4

Yuliya22 [3333]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

$R=\sqrt{P^2+Q^2+2PQCos\theta}$

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

$\theta= 90^o$

$R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70$

$(Cos90^o=0),(sin 90^o=1)$

$\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o$

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0

2 months ago

An automobile is traveling at a constant 15 m/s, then it undergoes acceleration from that moment forward. Which statement best d

ValentinkaMS [3465]

Answer:

d

Explanation:

7 0

1 month ago