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Ad libitum [116K]
3 years ago
9

Calculate the acceleration of a student riding his bicycle in a straight line that speeds up from 4 m/s to 6 m/s in 5 seconds.

Physics
1 answer:
solong [7]3 years ago
7 0
Acceleration is equal to velocity final minus velocity initial divided by time. 6m/s minus 4m/s divided by 5 seconds is 0.4m/s^2.

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During the spin cycle of a washing machine, the clothes stick to the outer wall of the barrel as it spins at a rate as high as 1
Darya [45]

To answer the two questions, we need to know two important equations involving centripetal movement:

v = ωr (ω represents angular velocity <u>in radians</u>)

a = \frac{v^{2}}{r}

Let's apply the first equation to question a:

v = ωr

v = ((1800*2π) / 60) * 0.26

Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.

Back to the equation:

v = ((1800*2π)/60) * 0.26

v = (1800*2(3.14)/60) * 0.26

v = (11304/60) * 0.26

v = 188.4 * 0.26

v = 48.984

v = 49 (m/s)

Now that we know the linear velocity, we can find the centripetal acceleration:

a = \frac{v^{2}}{r}

a = \frac{49^{2}}{0.26}

a = 9234.6 (m/s^{2})

Wow! That's fast!

<u>We now have our answers for a and b:</u>

a. 49 (m/s)

b. 9.2 * 10^{3} (m/s^{2})

If you have any questions on how I got to these answers, just ask!

- breezyツ

5 0
3 years ago
An inexperienced researcher runs an experiment and sets his alpha level at .40 because he can't wait to get his firstsignificant
GuDViN [60]

A potential problem is that you are willing to accept a <u>5% </u>chance of being wrong if you reject the null hypothesis.

The significance level is the probability of rejecting the null hypothesis if it is true. For example, a significance level of 0.05 indicates a 5% risk of concluding that there is a difference when there is actually no difference. Rejecting the true null hypothesis results in a Type I error.

The smaller the value of α the more difficult it is to reject the null hypothesis. Therefore, choosing a low value for α can reduce the likelihood of Type I errors. The result here is that if the null hypothesis is false, it may be more difficult to reject using a lower value for α. The alpha value or statistical significance threshold is arbitrary. Which value to use depends on your field of study.

Learn more about The potential problems here:-brainly.com/question/21836542

#SPJ4

4 0
1 year ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
In Hooke’s law, what does the x represent?
Arte-miy333 [17]
X Represents the distance the spring is stretched or compressed away from its equilibrium or rest position.
7 0
3 years ago
Read 2 more answers
Imagine you are holding an apple. how could you increase the potential energy of this apple?
Ray Of Light [21]

If you're holding the apple at your waist, lift it to your mouth.
Potential energy relative to any level is proportional to its height
above that level.  Increase that height, and you've increased the
potential energy.

Since energy is conserved ... it never magically appears or
disappears ... you need to tell where that extra energy for the
apple came from.

It's exactly the work you did ... the force of your muscles acting
through the distance you raised the apple ... that became the
additional potential energy that the apple gained.

6 0
3 years ago
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