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2 years ago

What was your train of thought as you navigated the picture of the candle?

Physics

1 answer:

defon2 years ago

4 0

Answer:

Where is the picture

Explanation:

WHERE IS THE PICTURE

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When an athlete holds a barbell overhead, the reaction force is the weight of the barbell on his hand. how does this force vary

Sonja [21]

The force applied by the competitor is littler than the heaviness of the barbell. At the point when the barbell quickens upward, the power applied by the competitor is more prominent than the heaviness of the barbell. When it decelerates upward, the power applied by the competitor is littler than the heaviness of the barbell.

3 0

3 years ago

Solve the question that follows using the equation for the conversion of Celsius to Fahrenheit. F=95(C)+32 On February 9, 1934,

SVEN [57.7K]

Answer:

Buffalo, NY

Explanation:

Temperature in Buffalo, NY = -29°C

In order to compare the temperatures we need to convert them to the same scale.

$F=\dfrac{9}{5}C+32\\\Rightarrow F=\dfrac{9}{5}\times -29+32\\\Rightarrow F=-20.2$

So, the temperature in Buffalo, NY was -20.2°F and the temperature in Anchorage, AL was 19°F.

Hence, it was colder in Buffalo, NY than in Anchorage, AL.

3 0

2 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago

State the law of conversation of energy in your own words. Give an example of a situation that you have either encountered or kn

RSB [31]

That Energy Cannot Be Created Nor Destroyed

4 0

2 years ago

When observing a specimen in a microscope you are told that the total magnification of the specimen is 630x assuming you are usi

Norma-Jean [14]

<span>Objective Lenses: Usually you will find 3 or 4 objective lenses on a microscope. They almost always consist of 4X, 10X, 40X and 100X powers. When coupled with a10X (most common) eyepiece lens, we get total magnifications of 40X (4X times10X), 100X , 400X and 1000X.</span>

5 0

3 years ago