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Find T1, the magnitude of the force of teTo practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.6 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.7 m from the other end. A monkey of mass 1.3 kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.nsion in string 1, at the moment that the monkey is halfway between the ends of the bar. Express your answer in newtons using three significant figures. View Available Hint(s)

Answers

Answer 1

Answer:

Answer:

$T_2 = 24.95 N$

$T_1 = 13.3 N$

Explanation:

As we know that total torque on the rod must be zero when monkey is at mid point of the rod

So we have

torque due to Tension at other end = torque due to weight of monkey + rod

so we will have

$T_2 (3 - 0.7) = (Mg + mg)(1.5)$

here we know that

M = 2.6 kg

m = 1.3 kg

$T_2(2.3) = (2.6 + 1.3)(9.81)(1.5)$

$T_2 = 24.95 N$

Now similarly we can say that

$T_1 + T_2 = (m + M)g$

$T_1 + 24.95 = (2.6 + 1.3)(9.81)$

$T_1 = 13.3 N$

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An object has an average acceleration of +6.07 m/s2 for 0.250 s . At the end of this time the object's velocity is +9.64 m/s .What was the object's initial velocity?

Answers

Answer:

Explanation:

From the question we are given;

Acceleration a = 6.07m/s²

Time t= 0.25s

Final velocity v = 9.64m/s

Required

Initial velocity u

Using the equation of motion

v = u+at

9.64 = u+(6.07)(0.25)

9.64 = u+1.5175

u = 9.64-1.5175

u = 8.1225m/s

Hence the object's initial velocity is 8.1225m/s

A rocket exhausts fuel with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with fuel comprising 80 per cent of the total mass. When all the fuel has been exhausted its speed is:________

Answers

The speed when all the fuel has been exhausted is 2415m/s

According to this question, the following information was given:

Exhaust velocity of fuel, V(e)= 1500 m/s
Initial speed of rocket, V₁ = 0 m/s
Final speed of rocket, V₂ = ?
Fuel weight = 80% of total weight

Using Tsiolkovsky rocket equation as follows:

∆V = V(e) ln(m1/m2)

m1 = initial mass
m2 = final mass without repellant

m2 = m1 - 80%m1

m2 = m1 - 0.8m1

m2 = 0.2m1

∆V = V2 - V1

Hence;

V2 - 0 = 1500 × ln (m1/0.2m1)

V2 = 1500 ln(1/0.2)

V2 = 1500 × 1.609

V2 = 2415m/s.

Therefore, the speed when all the fuel has been exhausted is 2415m/s.

Learn more at: brainly.com/question/19531823?referrer=searchResults

Answer:

$v_2 =2414\ m/s$

Explanation:

given,

exhaust velocity of fuel(v_e) = 1500 m/s

initial speed of rocket,v₁ = 0 m/s

final speed of rocket, v₂ = ?

fuel weigh = 80 % of total weight

using Tsiolkovsky rocket equation

$\Delta v = v_e ln((m_1)/(m_2))$

Δ v = v₂ - v₁

v_e is the exhaust speed

m₁ is the initial total mass.

m₂ is the is the final total mass without propellant.

m₂ = m₁ - 0.8 m₁

m₂ = 0.2 m₁

$v_2-v_1 = 1500* ln((m_1)/(0.2 m_1))$

$v_2 = 1500* ln((m_1)/(0.2 m_1))$

$v_2 =2414\ m/s$

When all the fuel is exhausted speed of the fuel is equal to $v_2 =2414\ m/s$

What minimum value of the coefficient of static friction between the ground and the cheetah's feet is necessary to provide this acceleration

Answers

Answer:

Coefficient of static friction = 1.84

Explanation:

Note:

Top speed = 60 mph

Acceleration of cheetah = 18 m/s²

Find:

Coefficient of static friction

Computation:

Acceleration due to gravity = 9.8 m/s²

Coefficient of static friction = Acceleration of cheetah / Acceleration due to gravity

Coefficient of static friction = 18 / 9.8

Coefficient of static friction = 1.84

A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be in- dependently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes 28 s to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular

Answers

Answer:

t = 39.60 s

Explanation:

Let's take a careful look at this interesting exercise.

In the first case the two motors apply the force in the same direction

F = m a₀

a₀ = F / m

with this acceleration it takes t = 28s to travel a distance, starting from rest

x = v₀ t + ½ a t²

x = ½ a₀ t²

t² = 2x / a₀

28² = 2x /a₀ (1)

in a second case the two motors apply perpendicular forces

we can analyze this situation as two independent movements, one in each direction

in the direction of axis a, there is a motor so its force is F/2

the acceleration on this axis is

a = F/2m

a = a₀ / 2

so if we use the distance equation

x = v₀ t + ½ a t²

as part of rest v₀ = 0

x = ½ (a₀ / 2) t²

let's clear the time

t² = (2x / a₀) 2

we substitute the let of equation 1

t² = 28² 2

t = 28 √2

t = 39.60 s

An emf is induced by rotating a 1207 turn, 20.0 cm diameter coil in the Earth's 4.13 10-5 T magnetic field. What average emf is induced, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 10.0 ms