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Mkey [24]
2 years ago
8

Question: For an 80-N squeeze on the handle of the pliers, determine the force F applied to the round rod b... For an 80-N squee

ze on the handle of the pliers, determine the force F applied to the round rod by each jaw. In addition, calculate the force supported by the pin at A.

Physics
1 answer:
balandron [24]2 years ago
6 0
<span>Force applied to round rod is 217 Newtons. Force applied to pin A is 297 Newtons. The pliers are basically a pair of simple levers. And like all levers, the product of length and force will remain a constant. So for the first part, we have the fulcrum at "A" The force applied is 80 N over a length of 95 mm giving us 80 * 95 = 7600. So the force applied to round rod will be X * 35 = 7600 And solving for X, gives us X * 35 = 7600 X = 217.1428571 N The answer to part 2 is almost as easy. But you need to rethink how the lever is. Instead of the fulcrum being at pin "A", we'll consider it to be at the round rod. So the lever arm that's getting the 80-N of force is now 35+95 = 130 mm long and the short level arm is still 35 mm. So 130 * 80 = 35 * X Solving for X 130 * 80 = 35 * X 10400 = 35 * X 297.1428571 = X So the pin is resisting 297 Newtons of force.</span>
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Susie walks 3 blocks north to the local CVS store, then 4 blocks east to her grandmother’s house. She then walks 2 blocks west a
Slav-nsk [51]

Answer:

Suzie is 3 blocks north of where she started

Explanation:

Displacement is the minimum distance between the initial and final point of motion.

Here, Suzie first walks 3 blocks north. From there she walks 4 blocks east. Then 2 blocks to the east then 2 blocks north and then 2 blocks east. She covered 4 blocks east toward west. This is the same distance she covered traveling east. But she is 2 blocks north. From there she traveled a block south to the pizzeria and another block to her friends house. She covered the two block she had traveled north.

Hence, Suzie is 3 blocks north of where she started.

7 0
2 years ago
If a galaxy is located 200 million light years from Earth, what can you conclude about the light from that galaxy?
natulia [17]
If a galaxy is located 200 million light years from Earth, you can conclude that t<span>he light will take 200 million years to reach Earth. </span>
8 0
2 years ago
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In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.
denis23 [38]
The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
5 0
2 years ago
Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get
olga_2 [115]

Answer:

Part a)

\omega = 8.17 rad/s

Part b)

N = 7.74 rev

Part c)

\alpha = 0.69 rad/s^2

Part d)

\alpha = 0.48 rad/s^2

Part e)

t = 9.14 s

Explanation:

Part a)

Angular speed is given as

\omega = 2\pi f

\omega = 2\pi(\frac{78}{60})

\omega = 8.17 rad/s

Part b)

Since turn table is accelerating uniformly

so we will have

\theta = \frac{\omega_f + \omega_i}{2} t

\theta = \frac{8.17 + 0}{2}(11.9)

2N\pi = 48.6

N = 7.74 rev

Part c)

angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{t}

\alpha = \frac{8.17 - 0}{11.9}

\alpha = 0.69 rad/s^2

Part d)

When its angular speed changes to 120 rpm

then we will have

\omega_2 = 2\pi (\frac{120}{60})

\omega_2 = 12.56 rad/s

number of turns revolved is 15 times

so we have

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)

\alpha = 0.48 rad/s^2

Part e)

now for uniform acceleration we have

\omega_f - \omega_i = \alpha t

12.56 - 8.17 = 0.48 t

t = 9.14 s

7 0
2 years ago
In Part 6.2.2, you will determine the wavelength of the laser by shining the laser beam on a "diffraction grating", a set of reg
harkovskaia [24]

Answer:

λ = 2042 nm

Explanation:

given data

screen distance d = 11 m

spot s = 4.5 cm = 4.5 ×10^{-2} m

separation L = 0.5 mm = 0.5 ×10^{-3} m

to find out

what is λ

solution

we will find first angle between first max and central bright

that is tan θ = s/d

tan θ = 4.5 ×10^{-2}  / 11

θ = 0.234

and we know diffraction grating for max

L sinθ  = mλ

here we know m = 1  so put all value and find λ

L sinθ  = mλ

0.5 ×10^{-3}  sin(0.234)  = 1 λ

λ = 2042.02 ×10^{-9}  m

λ = 2042 nm

3 0
2 years ago
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