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3 years ago

How is reflection of light used in research

1 answer:

7 0

In order to read the publications of his peers, or read his own notes of the work
that he did on the previous day, or find his coffee mug on his desk in the lab, the
research scientist must arrange to have each of them illuminated with visible
wavelengths of light, and then he must catch the light reflected from each of them
with his eyes.

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The gravitational force between two volleyball players is 3.9 × 10^-7 N . If the kasses of the players are 63 kg and 78 kg, what

N76 [4]

Answer:

I honestly have no idea bro.

Explanation:

Hope that helps have a great night.

3 0

2 years ago

The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (Th

Bezzdna [24]

Answer:

hello the diagram relating to this question is attached below

a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec

b) Force exerted on B2 at P = 39.2 N

Explanation:

Given data:

Co = 150 N-m ,

a) Determine the angular accelerations of B1 and B2 when couple is applied

at point P ; Co = I* ∝B2'

150 = ( (2*0.5^2) / 3 ) * ∝B2

∴ ∝B2' = 900 rad/sec

hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec

at point 0 ; Co = Inet * ∝B1

150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1

∴ ∝B1 = 180 rad/sec

hence angular acceleration of B1 = 180 rad/sec

b) Determine the force exerted on B2 at P

T2 = mB1g + T1 -------- ( 1 )

where ; T1 = mB2g ( at point p )

= 2 * 9.81 = 19.6 N

back to equation 1

T2 = (2 * 9.8 ) + 19.6 = 39.2 N

3 0

3 years ago

A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle

Vilka [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is $\alpha =2.538 \ rad/s^2$

Explanation:

From the question we are told that

The mass of the wheel is m = 6.9 kg

The radius is $r = 0.69 \ m$

The radius of gyration is $k_G = 0.4\ m$

The angle is $\theta = 47^o$

The force which the massless bar is subjected to $F = 22.5 \ N$

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that point A is the center of rotation.

Generally the moment of inertia about A is mathematically represented as

$I_a = I_G + M* r^2$

Here $I_G$ is the moment of inertia about G with respect to the radius of gyration which is mathematically represented as

$I_G = M * k_G$

=> $I_a = k_G* M + M* r^2$

=> $I_a =0.4 * 6.9 + 6.9 * 0.69^2$

=> $I_a =6.045 \ kg \cdot m^2$

Generally the torque experienced by the wheel is mathematically represented as

$\tau = F * cos (47)$

=> $\tau = 22.5 * cos (47)$

=> $\tau = 15.34 \ kg \cdot m^2 \cdot s^{-2}$

Generally this torque is also mathematically represented as

$\tau = I_a * \alpha$

=> $15.34 = 6.045 * \alpha$

=> $\alpha =2.538 \ rad/s^2$

4 0

3 years ago

A plane designed for vertical takeoff has a mass of 8.0 × 10³ kg. Find the net work done by all forces on the plane as it accele

artcher [175]

Answer:

the net work done after starting from rest is = 2.4 × 10⁵ J

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J), The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

Where m = mass of the plane, v = velocity of the plane, E = Energy, W = work done.

v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)

Substituting these values into equation 2

v² = 0² +2×1×30

v² = 60

v = √60

v = 7.75 m/s

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

Substituting these values into equation 1,

W = 1/2(8.0×10³)(7.75)²

W = (4.0×10³)(60)

W = 240 × 10³ J

W = 2.4 × 10⁵ J

Therefore the net work done after starting from rest is = 2.4 × 10⁵ J

4 0

3 years ago

A car, traveling at , encounters a dip in the road. The radius of curvature at the bottom of the dip is . Each of the car’s four

labwork [276]

Answer:

spring deflection is x = (v2 / R + g) m / 4

Explanation:

We will solve this problem with Newton's second law. Let's analyze the situation the car goes down a road and finds a dip (hollow) that we will assume that it has a circular shape in the lower part has the car weight, elastic force and a centripetal acceleration

Let's write the equations on the Y axis of this description

Fe - W = m $a_{c}$

Where Fe is elastic force, W the weight and $a_{c}$ the centripetal acceleration. The elastic force equation is

Fe = - k x

4 (k x) - mg = m v² / R

The four is because there are four springs, R is theradio of dip

We can calculate the deflection (x) of the springs

x = (m v2 / R + mg) / 4

x = (v2 / R + g) m / 4

5 0

3 years ago