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4 years ago

A 60 kg bicyclist going 2 m/s increased his work output by 1,800 J. What was his final velocity?

Physics

2 answers:

Alinara [238K]4 years ago

6 0

<em>8 </em>was the correct answer, thanks

Bas_tet [7]4 years ago

3 0

Bicyclist initial kinetic energy is Ek=(1/2)*m*v² where m is his mass and v is his speed and that is equal to:

Ek=(1/2)*60*2²=120 J.

When we add the increased work output, we get the total kinetic energy:

Ek(total)=Ek+W= 120 J + 1800 J= 1920 J

So Ek(total)=1920 J = (1/2)*m*V² where V is the speed after the bicyclist increased his work output. So lets solve for V:

(1/2)*60*V²=1920

30*V²=1920, we divide by 30,

V²=64, and take the square root of both sides,

V=8 m/s.

So the speed of the bicyclist after the increased work output is V=8 m/s.

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A coil has an inductance of 6.00 mH, and the current in it changes from 0.200 A to 1.50 A in a time interval of 0.300 s. Find th

Andre45 [30]

Answer:

0.026 V

Explanation:

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3 years ago

The block shown above has a mass of 105 g.<br> What is the density of the block? SC.8.P.8.3

zaharov [31]

The density of the block is 1.25 cm³

The correct answer to the question is Option B. 1.25 cm³

To solve this question, we'll begin by calculating the volume of the block. This can be obtained as follow:

Length = 7 cm

Height = 4 cm

Width = 3 cm

<h3>Volume =? </h3>

Volume = Length × Width × Height

Volume = 7 × 3 × 4

<h3>Volume = 84 cm³</h3>

Thus, the volume of the block is 84 cm³

Finally, we shall determine the density of the block. This can be obtained as follow:

Density is defined as mass per unit volume i.e

$Density = \frac{mass}{volume} \\\\$

Mass of block = 105 g

Volume of block = 84 cm³

<h3>Density of block =? </h3>

$Density = \frac{mass}{volume}\\\\Density = \frac{105}{84}\\\\$

<h3>Density of block = 1.25 cm³</h3>

Therefore, the density of the block is 1.25 cm³.

Hence, Option B. 1.25 cm³ gives the correct answer to the question.

Learn more: brainly.com/question/2040396?referrer=searchResults

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2 years ago

A 13.0 g piece of Styrofoam carries a net charge of â5.40 Î¼C and floats above the center of a large horizontal sheet of plastic

kati45 [8]

Answer:

σ =4.180×10^{-9} C/m^2

Explanation:

electric field due to non conducting sheet is

$E=\frac{\sigma}{2\epsilon_0}$

the force acting on the piece of Styrofoam

Eq= mg

⇒E= mg/q

now,

$\frac{mg}{q} = \frac{\sigma}{2\epsilon_0}$

⇒ $\sigma= \frac{2mg\epsilon_0}{q}$

$\sigma= \frac{2\times0.013\times9.81\times8.85\times10^{-12}}{5.40\times10^{-4}}$

charge per unit area (in C/m2) on the plastic sheet σ =4.180×10^{-9} C/m^2

6 0

3 years ago

A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta

yuradex [85]

Answer:

1.25 kgm²/sec

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Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

Disk angular speed =

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Ekd = 0.148 * 9.87

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Ekd = (Jc + 1/2*Jd) * ω²

Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

ω² = 1.4607/0.74

ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

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