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2 years ago

Determine the cutting force f exerted on the rod s in terms of the forces p applied to the handles of the heavy-duty cutter.

2 answers:

6 0

CASE 1: Direction vertically downward on the blade

Assume pin force at E acts downward on the blade and CW is a positive moment 
Ey[3.4] - F[1.7] = 0 
Ey = F/2 

Case 2: Direction vertically upward on the blade . Assume CW is positive moment 

Ey[1.5sin19] – P[21 – 1.5sin19] = 0 
(F/2)[1.5sin19] = P[21 – 1.5sin19] 
F = 2P[21 – 1.5sin19] / [1.5sin19] 
F = 84P

Alexus [3.1K]2 years ago

5 0

If you use the next formula with the data given in the exercise you are asking:
Ey[3.4] - F[1.7] = 0
Ey = F/2
and after that what you need to do is sum the moments of the handle about D to zero asumming it is a positive moment and we proceed like this
Ey[1.5sin19] – P[21 – 1.5sin19] = 0
(F/2)[1.5sin19] = P[21 – 1.5sin19] 
F = 2P[21 – 1.5sin19] / [1.5sin19] 
F = 84P

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Find the centripetal force needed by a 1275 kg car to make a turn of radius 40.0 m at a speed of 25.0 km/h

natima [27]

v = 25.0 km/h = 25*5/18 m/s = 6.94 m/s

centripetal force = mv²/r = 1275*6.94²/40 = 1537.18 N

4 0

2 years ago

Read 2 more answers

On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs accou

BabaBlast [244]

Answer:

176.38 rpm

Explanation:

mass percentage of arms and legs = 13%

mass percentage of legs and trunk = 80%

mass percentage of head = 7%

Total mass of the skater = 74.0 kg

length of arms = 70 cm = 0.7 m

height of skater = 1.8 m

diameter of trunk = 35 cm = 0.35 m

Initial angular momentum = 68 rpm

We assume:

The spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally.
friction between the skater and the ice is negligible.

We split her body into two systems, the spinning hands as spinning rods

1. Each rod has moment of inertia = $\frac{1}{3} mL^{2}$

mass m of the arms is 13% of 74 kg = 0.13 x 74 = 9.62 kg

mass of each side will be assumed to be 9.62/2 = 4.81 kg

L = length of each arm

therefore,

I = $\frac{1}{3}$ x 4.81 x $0.7^{2}$ = 0.79 kg-m for each arm

2. Her body as a cylinder has moment of inertia = $\frac{1}{2} mr^{2}$

r = radius of her body = diameter/2 = 0.35/2 = 0.175 m

mass of body trunk = (80% + 7%) of 74 kg = 0.87 x 74 = 64.38 kg

I = $\frac{1}{2}$ x 64.38 x $0.175^{2}$ = 0.99 kg-m

We consider each case

case 1: Body spinning with arm outstretched

Total moment of inertia = sum of moments of inertia of both arms and moment of inertia of body trunk

I = (0.79 x 2) + 0.99 = 2.57 kg-m

angular momentum = Iω

where ω = angular speed = 68.0 rpm = $\frac{2\pi }{60}$ x 68 = 7.12 rad/s

angular momentum = 2.57 x 7.12 = 18.29 kg-rad/m-s

case 2: Arms pulled down parallel to trunk

The momentum of inertia will be due to her body trunk alone which is 0.91 kg-m

angular momentum = Iω

= 0.99 x ω = 0.91ω

according to conservation of angular momentum, both angular momentum must be equal, therefore,

18.29 = 0.99ω

ω = 18.29/0.99 = 18.47 rad/s

18.47 ÷ $\frac{2\pi }{60}$ = 176.38 rpm

7 0

2 years ago

Energy is observed in two basic forms: potential and kinetic. Which of the following correctly matches these forms with a source

Travka [436]

Answer:

option C

Explanation:

The correct answer is option C

Kinetic energy is the energy which is due to the motion of body.

Potential energy is the energy due to virtue of position of the object.

option A is not true because potential energy is due the position of the body

Option B should be the potential energy not kinetic energy.;

Option D is motion of individual molecule leads to kinetic energy not potential energy.

So, the correct answer is option is the covalent bonds of a sugar molecule is potential energy because of the position of bond.

4 0

2 years ago

Each metal is illuminated with 400 nm (3.10 eV) light. Rank the metals on the basis of the maximum kinetic energy of the emitted

34kurt

Answer:

K.E(K) > K.E(Cs) > 0 (others)

Explanation:

Given the Work functions of the metal as

Aluminium (Wo)=4eV

Platinum(Wo) =6.4eV

Cesium (Wo) =2.1eV

Beryllium (Wo) = 5.0eV

Magnesium (Wo) = 3.7eV

Potassium (Wo) = 2.3eV

Using the formula:

K.E = hf - Wo........(1)

Wo = hfo..............(2)

From these the fo can be calculated for all the metals

Where K.E =Kinetic Energy

hf = energy of illumination = 3.10eV

h is Planck constant and has the value 6.6 × 10^-34JS^-1

The frequency f of the illumination is given by

f = 3.10 × 1.6 × 10^-19/6.6 × 10^-34

f = 7.51 × 10¹⁴ Hz..........(*)

Now an electron is only ejected if the threshold frequency of the metal is reached.

The work function has a threshold frequency (fo) for all the metals and this minimum frequency required to required to remove an electron from the surface of a metal.

We need to compare f with fo

If fo >= f there is emission, otherwise there is no emission

So using (2) we calculate for all fo and compare with f

K.E(Al) = 3.10 - 4.0 - 3.10 = -0.9eV, fo = 9.70 × 10¹⁴ Hz (no emission)

K.E(Pt) = 3.10 - 6.40 = -3.30eV, fo = 1.55 × 10^15 Hz, ( no emission)

K.E(Cs) = 3.10 - 2.10 = -1.0eV, fo = 5.09×10¹⁴ Hz, (emission)

K.E(Be) =3.10-5.0 = -1.90eV, fo = 12.12 ×10^15 Hz.,(no emission)

K.E(Mg) = 3.10-3.70 = -0.6eV, fo = 8.97 × 10¹⁴Hz, (no emission)

K.E(K) = 3.10 - 2.30= 0.9eV, fo = 5.58 × 10¹⁴ Hz, (emission)

So the metals whose electron gain Kinetic energy are:

Cesium

Potassium

Others have zero kinetic energy since no electron is emitted.

Hence the rank is:

K.E(K) > K.E(Cs) > 0 (others)

6 0

2 years ago

Two narrow, parallel slits separated by 0.85 mm are illuminated by 600 nm light, and the viewing screen is 2.8 m away from the s

AURORKA [14]

Answer:

Phase difference = pi/4 radians

Explanation:

Given:

- The wavelength of incident light λ = 600 nm

- The split separation d = 0.85 mm

- Distance of screen from split plane L = 2.8 m

Find:

What is the phase difference between the two interfering waves on a screen, at a point 2.5 mm from the central bright fringe?

Solution:

- The phase difference can be evaluated by determining the type of interference that occurs at point y = 2.5 mm above central order. We will use the derived results from Young's double slit experiment.

sin ( Q ) = m*λ /d

m = d*sin(Q) / λ

- Where, m is the order number and angle Q is the angle for mth order of fringe from central bright fringe.

r = sqrt ( L^2 + 0.0025^ )

Where, r is the distance from split to the interference bright fringe.

r = sqrt(2.8^ + 0.0025^) = 2.8

sin(Q) = 0.0025 / 2.8

Hence. m = 0.00085*0.0025 / 2.8*(600*10^-9)

m = 1.26

- We know that constructive interference would occurred at m = 1 and destructive interference @ m = 1.5. They have a phase difference of pi/2 radians.

- The order number lies in between constructive and destructive interference i.e m ≈ 1.25 then the corresponding phase difference = 0.5*(pi/2).

Answer: Phase difference = pi/4 radians

6 0

2 years ago