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2 years ago

13

in a race, Usain Bolt accelerates at 1.99m/s^2 for the first 60.0m, then decelerates at -0.266m/s^2 for the final 40.0m. what wa

s his final velocity? (Unit = m/s)

1 answer:

Paul [167]2 years ago

6 0

Answer:

14.7 m/s

Explanation:

During the first phase, given:

Δx = 60.0 m

v₀ = 0 m/s

a = 1.99 ms²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (1.99 m/s²) (60.0 m)

v = 15.5 m/s

During the second phase, given:

Δx = 40.0 m

v₀ = 15.5 m/s

a = -0.266 ms²

Find: v

v² = v₀² + 2aΔx

v² = (15.5 m/s)² + 2 (-0.266 m/s²) (40.0 m)

v = 14.7 m/s

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A weightlifter lifts a 13.0-kg barbel from the ground an moves it a distance of 1.3 meters. What is the work se does on the barb

marta [7]

The work done on the barbell is -165.62 Nm.

Explanation:

Work done on any object is the measure of force required to move that object from one position to another. So it is determined by the product of force acting on the object with the displacement of the object.

In the present problem, the displacement of the object on acting of force is given as 1.3 m. And the weight of the object which is a barbel is given as 13 kg. As the work is to lift the object from the ground, so the acceleration due to gravity will be acting on the object. In other words, the force applied on the object to lift it should be in opposite direction to the acting of acceleration due to gravity.

Thus, $Force = - Mass * Acceleration due to gravity = - 13 * 9.8 =-127.4 N$

Now, the force is -127.4 N and the displacement is 1.3 m.

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2 years ago

A couch is pushed with a horizontal force of 80 N and moves the couch a

Lapatulllka [165]

Answer:

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2 years ago

Read 2 more answers

A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th

expeople1 [14]

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

Conduction could be described by the Law of Fourierin the form: $Q=kA\frac{T_1-T_2}{L}$ where $Q$ is the rate of heat transferred by conduction, $k$ is the thermal conductivity of the material, $T_1$ and $T_2$ are the temperatures of each heat deposit, $A$ is the cross area to the flow of heat, and ${L}$ is the distance that the flow of heat has to go.
For the original cylinder the Fourier's law would be: $kA_1\frac{T_1-T_2}{L_1}=25W$ , and if $A_1=\frac{\pi D_{1}^{2}}{4}$ , then the expression would be: $k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W$ where $D_1$ is the diameter of the original cylinder, and ${L_1}$ is the length of the original cylinder.
For the new cylinder, in the same fashion that for the first, Fourier's Law would be: $Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2}$ ,where $Q_2$ is the heat rate in the second case, $D_2$ and ${L_2$ are the new diameter and length.
But, $D_2=2D_1$ and $L_2=2L_1$ , substituting in the expression for $Q_2$ : $Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}$ .
Rearranging: $Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1})$ .
In the last declaration of $Q_2$ , it could be noted that the expressión inside the parenthesis is actually $Q_1$ , then: $Q_2=\frac{2^2}{2}(25W)=50W$ .
<u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>

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2 years ago

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tankabanditka [31]

Answer

given,

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so, from the diagram attached below

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$F_b cos {\theta_b} = F_L sin {\theta_L}$

$cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}$

$cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}$

$cos {\theta_b} = 0.464$

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2 years ago

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andrey2020 [161]

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2 years ago