An object moving with 108 km/h moves 400 m in 8 seconds. find the velocity attained by the object.

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Hunter-Best [27]

Answer:

um how about no.. this is not the site for what you're looking for...

Explanation:

4 0

3 years ago

Read 2 more answers

A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in

lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

A = $1.5 \times 2.30$

= 3.45 $in^{2}$

The given data is as follows.

Allowable stress = 14,500 psi

Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

$P_{max} = \sigma_{a}A$

= $14500 \times 3.45$

= 50025 lb

Now, maximum load from shear stress is as follows.

$P_{max} = 2 \times \tau_{a} \times A$

= $2 \times 7100 \times 3.45$

= 48990 lb

Hence, $P_{max}$ will be calculated as follows.

$P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})$

= 48990 lb

Thus, we can conclude that the maximum permissible load $P_{max}$ is 48990 lb.

4 0

3 years ago

Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the pow

allsm [11]

This question is incomplete, the complete question;

you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.

Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity

Options;

a) 200 nm; 0.9 mW

b) 100 nm, 0.0059 mW

c) 200 nm; 0 mW

d) 100 nm; 0.9 mW

e) 200 nm; 0.0059 mW

Answer:

the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer

Explanation:

Given that the instrument here is an interferometer.

Maximum intensity is obtained when the two waves are exactly in phase.

that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.

The phase factor of this point is taken as ∅ = 0

Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π

This corresponds to a path difference between the two waves as half of the wavelength. λ/2

The light gets reflected from the mirror.

Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)

hence, to get a path difference of λ/2 the mirror should move half of this distance only

so, the mirror should move;

$l$ = λ/4

here, wavelength is 400nm

the length moved by the mirror = 400/4 = 100 nm

The intensity is given by the equation;

$l$ = $l$ 1 + $l$ 2 + 2√ $l$ 1 $l$ 2cos(∅)

where

$l$ 1 = 2.25 mW

$l$ 2 = 2.025 mW

∅ = π

so we substitute

$l$ = 2.25 + 2.025 - 2√(2.25 × 2.025)

$l$ = 4.275 - 4.26907

$l$ = 0.0059

Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;

100 nm; 0.0059 mW

Option b) 100 nm, 0.0059 mW is the correct answer

5 0

2 years ago

A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha

Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

$N = \dfrac{m g}{cos\theta-\mu sin \theta}$

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

$\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}$

taking cos θ from nominator and denominator

$F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg$

$\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg$

$\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g$

$\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}$

now, inserting all the given values

$\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}$

μ = 0.6

7 0

3 years ago

A vector A has a magnitude of 5 units and points in the −y-direction, while a vector B has triple the magnitude of A and points

Harman [31]

Answer:

A+B; 5√5 units, 341.57°

A-B; 5√5 units, 198.43°

B-A; 5√5 units, 18.43°

Explanation:

Given A = 5 units

By vector notation and the axis of A, it is represented as -5j

B = 3 × 5 = 15 units

Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B

(a) A + B = -5j + 15i

A+B = 15i -5j

|A+B| = √(15)²+(5)²

= 5√5 units

∆ = arctan(5/15) = 18.43°

The angle ∆ is generally used in the diagrams

∆= 18.43°

The direction of A+B is 341.57° based in the condition given (see attachment for diagrams

(b) A - B = -5j -15i

A-B = -15i -5j

|A-B|= √(15)²+(-5)²

|A-B| = √125

|A-B| = 5√5 units

The direction is 180+18.43°= 198.43°

See attachment for diagrams

(c) B-A = 15i -( -5j) = 15i + 5j

|B-A| = 5√5 units

The direction is 18.43°

See attachment for diagram

5 0

3 years ago