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WINSTONCH [101]

3 years ago

5

forces F1 (east) and F2 are simultaneously applied to a 3.0kg mass. when F2 is east, a=5.0m/s² east and when F2 is west a=1.0m/s

² east. Determine the magnitude of F1 and F2.

1 answer:

zhenek [66]3 years ago

5 0

Answer:

345453-5676

its the right answer

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A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.070

likoan [24]

Given Information:

length of slender rod = L = 90 cm = 0.90 m

mass of slender rod = m = 0.120 kg

mass of sphere welded to one end = m₁ = 0.0200 kg

mass of sphere welded to another end = m₂ = 0.0700 kg (typing error in the question it must be 0.0500 kg as given at the end of the question)

Required Information:

Linear speed of the 0.0500 kg sphere = v = ?

Answer:

Linear speed of the 0.0500 kg sphere = 1.55 m/s

Explanation:

The velocity of the sphere can by calculated using

ΔKE = ½Iω²

Where I is the moment of inertia of the whole setup ω is the speed and ΔKE is the change in kinetic energy

The moment of inertia of a rigid rod about center is given by

I = (1/12)mL²

The moment of inertia due to m₁ and m₂ is

I = (m₁+m₂)(L/2)²

L/2 means that the spheres are welded at both ends of slender rod whose length is L.

The overall moment of inertia becomes

I = (1/12)mL² + (m₁+m₂)(L/2)²

I = (1/12)0.120*(0.90)² + (0.0200+0.0500)(0.90/2)²

I = 0.0081 + 0.01417

I = 0.02227 kg.m²

The change in the potential energy is given by

ΔPE = m₁gh₁ + m₂gh₂

Where h₁ and h₂ are half of the length of slender rod

L/2 = 0.90/2 = 0.45 m

ΔPE = 0.0200*9.8*0.45 + 0.0500*9.8*-0.45

The negative sign is due to the fact that that m₂ is heavy and it would fall and the other sphere m₁ is lighter and it would will rise.

ΔPE = -0.1323 J

This potential energy is then converted into kinetic energy therefore,

ΔKE = ½Iω²

0.1323 = ½(0.02227)ω²

ω² = (2*0.1323)/0.02227

ω = √(2*0.1323)/0.02227

ω = 3.45 rad/s

The linear speed is

v = (L/2)ω

v = (0.90/2)*3.45

v = 1.55 m/s

Therefore, the linear speed of the 0.0500 kg sphere as its passes through its lowest point is 1.55 m/s.

8 0

3 years ago

A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If

inessss [21]

Answer:

the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

v² = v₀² + 2 a₁ x

indicate that the initial velocity is zero

v² = 2 a₁ x

let's calculate

v = $\sqrt {2 \ 15.0 \ 645}$

v = 143.666 m / s

now for the second interval let's find the distance it takes to stop

v₂² = v² - 2 a₂ x₂

in this part the final velocity is zero (v₂ = 0)

0 = v² - 2 a₂ x₂

x₂ = v² / 2a₂

let's calculate

x₂ = $\frac{ 143.666^2 }{2 \ 18.2}$

x₂ = 573 m

as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

3 0

3 years ago

A lamp is labelled '230 V, 100 W'. How many joules of electrical energy is changed to thermal energy and light if the lamp is sw

viva [34]

Given;

V = 230V

Power, P = 100W

time, t = 2hrs = 7200s

from,

P = IV

and Energy, E= Pt

E = 100*7200

E = 720000 Joules

E = 720KJ

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Sveta_85 [38]

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Reptile [31]

Answer:

speed

Explanation:

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