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S_A_V [24]
3 years ago
14

Two events are observed in a frame of reference S to occur at the same space point, with the second event occurring after a time

of 1.70 s . In a second frame S' moving relative to S, the second event is observed to occur after a time of 2.25 s . What is the difference between the positions of the two events as measured in S'?

Physics
1 answer:
Deffense [45]3 years ago
5 0

Answer:

he difference between the positions of the two events as measured in S = 4.42 x 10 8 m

Explanation:

The concept of time dilation is applied to solve the question as shown in the attachment.

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A particle moving on a plane curve what is the degree of freedom
kirill115 [55]

A particle confined to move along a curved path has only one degree of freedom. inclined plane are some examples of constrained motion. Every condition of constraint reduces the number of degree of freedom by one.

I hope this helps!

7 0
3 years ago
A child sits 1.6 m from the center of a merry go round that makes one complete revolution in 4.7 s. What is his angular accelera
olga2289 [7]

Answer: angular acceleration = 2.86\ rad/sec^{2}

Given:

Distance from center of axis = 1.6 m

Time taken to complete one revolution = 4.7 sec

Therefore, we can evaluate the angular acceleration using the following formula:

\alpha = r\times \omega^{2}

\alpha = r\times (\frac{2\pi}{T})^{2}

\alpha = 1.6\times (\frac{2\times3.14}{4.7})^{2}

\alpha = 2.86\ rad/sec^{2}

7 0
3 years ago
A helicopter's speed increases
sergij07 [2.7K]

Answer:

the helis speed increased its speed by 35 m/s. Unless you are asking me what was the speed per second. if you were asking me for the speed per second then that would be a different answer

Explanation:

7 0
3 years ago
Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th
alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box m=3\ kg

Force applied is F=25\ N

Displacement of the box is s=15\ m

Velocity acquired by the box is v=6\ m/s

acceleration associated with it is a=\dfrac{F}{m}

\Rightarrow a=\dfrac{25}{3}\ m/s^2

Work done by force is W=F\cdot s

W=25\times 15\\W=375\ J

change in kinetic energy is \Delta K

\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J

Therefore, the magnitude of work done by friction is 321\ J

3 0
3 years ago
Given the scope of Kayla’s education, training, and years of experience as a CMA, would this “favor” fall within the AAMA guidel
Cerrena [4.2K]

Based on the information given, it can be inferred that the favor doesn't fall within the AAMA guidelines of her responsibilities.

From the information given, it should be noted that the guidelines of CMA as stipulated under the American Association of Medical Assistant prohibits the CMA from interpreting the medical data of the patient. Therefore, the favor that was asked by Dr. Hsu of Kayla is simply against the guidelines.

Even though the favor that was asked by Dr. Hsu was prohibited by AAMA, it should be noted that the final part of the favor about faxing the report to the internist would fall within AAMA guidelines.

In conclusion, the best way that Kayla can respond to Dr. Hsu is to decline doing the favor.

Read related link on:

6 0
3 years ago
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