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3 years ago

How much work must be done on a 10-kg bicycle to increase its speed from 5

Physics

1 answer:

Karo-lina-s [1.5K]3 years ago

6 0

Answer:

375 j

Explanation:

Work done = increase in kinetic energy

C is the correct answer for this question. 375 work must be done on a 10 kg bicycle.

1: 1/5*10 * (10+5)(10-5)

Work Done: 375J

<em><u>Hope this helps.</u></em>

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The formula $F = \frac{9}{5} C + 32$ can be used to convert temperatures between degrees Fahrenheit ($F$) and degrees Celsius ($

slamgirl [31]

Answer:

-30° C

Explanation:

Data provided in the problem:

The formula for conversion as:

F = (9/5)C + 32

Now,

for the values of F = -22 , C = ?

Substituting the value of F in the above formula, we get

-22 = (9/5)C + 32

-22 - 32 = (9/5)C

(9/5)C = - 54

C = - 54 × (5/9)

C = - 30 °

Hence, -22 Fahrenheit equals to -30°C

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3 years ago

The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write this is powers of ten i

DerKrebs [107]

1.4 times 10 in power of 10

8 0

3 years ago

An object with a non-zero speed must be _________.

pogonyaev

It would be either A or C if its still moving and not stopping

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3 years ago

Read 2 more answers

Find the gravitational potential energy of an 84 kg person standing atop Mt. Everest at an altitude of 8848 m. Use sea level as

djverab [1.8K]

Answer:

$E=7.28\times 10^6\ J$

Explanation:

Given that,

Mass of a person, m = 84 kg

The person is standing at a top of Mt. Everest at an altitude of 8848 m

We need to find the gravitational potential energy of the person. We know that the gravitational potential energy is possessed due to the position of an object. It is given by :

E = mgh, g is the acceleration due to gravity

$E=84\ kg\times 9.8\ m/s^2\times 8848\ m\\\\E=7283673.6\ J\\\\E=7.28\times 10^6\ J$

So, the gravitational potential energy of the person is $7.28\times 10^6\ J$

6 0

3 years ago

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0

3 years ago