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2 years ago

What’s a real life example of relative motion?

Physics

1 answer:

vesna_86 [32]2 years ago

7 0

Answer:

let's say you're on a bus going 50 km/hr, you are moving at a velocity of zero relative to the bus. however, relative to the ground you are moving at the same velocity as the bus.

Explanation:

physics

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A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

3 years ago

Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the

Gnesinka [82]

Answer:

F = 233.52 N, θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

= 80.5 N

Jill

cos 45 = F_{2x} / F₂2

sin 45 = F_{2y} / F₂2

F_{2x} = F₂ cos 45

F_{2y} = F₂ sin 45

F_{2x} = 81.7 cos 45 = 57.77 N

F_{2y} = 81.7 sin 45 = 57.77 N

Jane

cos (270 + 45) = F_{3x} / F₃3

sin 315 = F_{3y} / F₃

F_{3x} = 131 cos 315 = 92.63 N

F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

F_{x} = F_{1x} + F_{2x} + F_{3x}

F_{x} = 80.5 +57.77 + 92.63

F_{x} = 230.9 N

Axis y

F_{y} = F_{1y} + F_{2y} + F_{3y}

F_{y} = 0 + 57.77 -92.63

F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

F = √(Fₓ² + F_{y}²

F = √(230.9² + 34.86²)

F = 233.52 N

let's use trigonometry for the angle

tan θ = $\frac{F_y}{F_x} }$

θ = tan⁻¹ (\frac{F_y}{F_x} })

θ = tan⁻¹ (-34.86 / 230.9)

θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

θ' = 360 -θ

θ‘= 360- 8.59

θ' = 351.41º

5 0

3 years ago

Which state of matter has the greatest distance between the individual particles?

Maurinko [17]

Your answer is c steam because steam is a gas...

4 0

3 years ago

Read 2 more answers

Is god real??? i wanna know

NemiM [27]

In my opinion, yes the bible tell us that "For God so loved the world that he gave his one and only Son, that whoever believes in him shall not perish but have eternal life"
So my answer is yes

6 0

3 years ago

Read 2 more answers

I''l give brainliest. Please help. I hold 2 objects about 0.1 meters apart. What is the electrostatic force between the two obje

matrenka [14]

Answer:

F = K Q1 Q2 / R^2 where K = 9 * 10E9 (1 / 4 pi ∈0)

F = 9.00E9 * (4.6E-16)^2 / .01 = 1.90E-19 N

7 0

2 years ago