Ask question

Ask question

All categories

2 years ago

12

Two wheels can rotate freely about fixed axles through their centers. The wheels have the same mass, but one has twice the radiu

s of the other. Forces F1 and F2 are applied as shown such that the angular acceleration of the two wheels is the same.
1)Compare the magnitudes of the two forces
F2 = F1
F2 = 2 F1
F2 = 4 F1

1 answer:

Murrr4er [49]2 years ago

7 0

Answer:

The force on second wheel is twice off the force on first wheel.

Explanation:

In this case, two wheels can rotate freely about fixed axles through their centers. We know that, in rotational mechanics, the torque is given by :

$\tau=I\alpha$

Also, $\tau=Fr$

And moment of inertia is, $I=mr^2$

It implies,

$Fr=mr^2\alpha \\\\F=mr\alpha$

Here, one has twice the radius of the other. Ratio of forces will be :

$\dfrac{F_1}{F_2}=\dfrac{mr_1\alpha }{mr_2\alpha }\\\\\dfrac{F_1}{F_2}=\dfrac{r_1 }{2r_1}\\\\\dfrac{F_1}{F_2}=\dfrac{1 }{2}\\\\F_2=2F_1$

So, the force on second wheel is twice off the force on first wheel.

You might be interested in

You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.

Andru [333]

Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.

v = v0 + g · t

6.00 m/s = 12.0 m/s -9.81 m/s² · t

(6.00 - 12.0)m/s / -9.81 m/s² = t

t = 0.612 s

Now, let´s calculate the height of the baseball at that time:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²

y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Have a nice day!

4 0

2 years ago

An inline skater skates on a circular track 120.0 m in diameter at a tangential speed of 9.20 m/s. If the skater’s mass is 68.5

jok3333 [9.3K]

Answer:

The centripetal force acting on the skater is <u>48.32 N.</u>

Explanation:

Given:

Radius of circular track is, $R=120.0\ m$

Tangential speed of the skater is, $v=9.20\ m/s$

Mass of the skater is, $m=68.5\ kg$

We are asked to find the centripetal force acting on the skater.

We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.

Centripetal force acting on the skater is given as:

$F_c=\frac{mv^2}{R}$

Now, plug in the given values of the known quantities and solve for centripetal force, $F_c$ . This gives,

$F_c=\frac{68.5\times (9.20)^2}{120.0}\\\\F_c=\frac{68.5\times 84.64}{120}\\\\F_c=\frac{5797.84}{120}\\\\F_c=48.32\ N$

Therefore, the centripetal force acting on the skater is 48.32 N.

3 0

2 years ago

A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high- efficiency moto

daser333 [38]

Convert the shaft ouput from HP to kW

Shaft output = 75HP = 55.93kW

1st: Finding for the power consumption based on 55.93kW output

Power consumption (Old) = 55.93kW / .910 = 61.46kW

Power consumption (New) = 55.93kW / .954 = 58.63kW

2nd: Total power used in kWh:

Power Used = Power consumption * load factor * Hours:

Power (Old) = 61.46kW * .75 * 4368 = 201343 kWh

Power (New) = 58.63kW * .75 * 4368 = 192072 kWh

Energy saved = 201343 kWh - 192072 kWh = 9,271 kWh

3rd: Calculating for the price:

Price = kW-Hr * $/kWh

Price (Old) = 201343kWh * $0.08/kWh = $16107.44

Price (New) = 192072 kWh * $0.08/kWh = $15365.76

Cost saved = $16107.44 - $15365.76 = $741.68/yr

4th: Setting up the cost equation:

Cost over time, F(t) = Motor_Cost + (Price * Number of Years, t)

Cost (Old) = 5449 + 16107.44*t

Cost (New) = 5520 + 15365.76*t

Equate the two to find for t when they cost equally:

5449 + 16107.44*t = 5520 + 15365.76*t

16107.44*t = 15365.76*t +71

16107.44*t - 15365.76*t = 71

741.68*t = 71

t = 71 / 741.68 = .095 years = 35 days

So the payback period is after 35 days.

6 0

2 years ago

When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant

weqwewe [10]

Answer:

a) v = 1.19 m / s , b) P₁ = 0.922 10⁵ Pa

Explanation:

1) Let's use the fluid continuity equation

Q = A v

The area of a circle is

A = π r2 = π d²/4

v = Q / A = Q 4 / pi d²

v = 0.006 4/π 0.08²

v = 1.19 m / s

2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point

P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂

The exercise tell us

P₂ = 1.0013 105 Pa

v₁ = 0

y₁ = 1 m

y₂=0

Rho (water) = 1000 kg / m³

P₁ + rho y₁ = P₂ + ½ rho v₂²

P₁ = P₂ + ½ rho v₂² - rho g y₁

P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

P₁ = 1.013 10⁵ +708.5 - 9800

P₁ = 92208.5Pa

P₁ = 0.922 10⁵ Pa

8 0

2 years ago

A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the

likoan [24]

Answer:

The separation between the first two minima on either side is 0.63 degrees.

Explanation:

A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:

$a\sin \theta_n=n\lambda$

with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:

for the first minimum

$a\sin \theta_1=(1)\lambda$

solving for θ1:

$\theta_1=\arcsin (\frac{\lambda}{a})=\arcsin (\frac{550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_1=0.63 degrees$

for the second minimum:

$a\sin \theta_2=(2)\lambda$

$\theta_2=\arcsin (\frac{2\lambda}{a})=\arcsin (\frac{2*550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_2=1.26 degrees$

So, the angular separation between them is the rest:

$\Delta \theta =1.26-0.63$

$\Delta \theta=0.63$

4 0

2 years ago