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3 years ago

An apple contains 165 Calories. How many actual calories does it contain? How many joules does it contain

Physics

1 answer:

jasenka [17]3 years ago

8 0

Answer:

95 calories

Explanation:

A medium-sized apple has only 95 calories but plenty of water and fiber.

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Which moon has a thick atmosphere made mostly of nitrogen?

Delicious77 [7]

the answer is Titan
explanation: titan fascinates scientists because of its thick atmosphere — which is mostly made of nitrogen gas

7 0

3 years ago

Find the first three harmonics of a string of linear mass density 2. 00 g/m and length 0. 600 m when the tension in it is 50. 0

gavmur [86]

The first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

<h3>Velocity of the wave</h3>

The velocity of the wave is calculated as follows;

v = √T/μ

where;

T is tension
μ is mass per unit length = 2 g/m = 0.002 kg/m

v = √(50/0.002)

v = 158.1 m/s

<h3>First harmonic or fundamental frequency of the wave</h3>

f₀ = v/λ

where;

λ is the wavelength = 2L

f₀ = v/2L

f₀ = 158.1/(2 x 0.6)

f₀ = 131.8 Hz

<h3>Second harmonic of the wave</h3>

f₁ = 2f₀

f₁ = 2(131.8 Hz)

f₁ = 263.6 Hz

<h3>Third harmonic of the wave</h3>

f₂ = 3f₀

f₂ = 3(131.8 Hz)

f₂ = 395.4 Hz

Thus, the first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

Learn more about harmonics here: brainly.com/question/4290297

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6 0

1 year ago

Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and

djverab [1.8K]

The magnitude of the average friction force exerted on the collar (F)= 8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

$\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c} = mgy$

Here we are given,

k = The spring has a spring constant.

= 25.5 N/m.

$x_f$ = Final length of the spring .

= $\sqrt{1.25^2+1.8^2} -0.60$

= 1.591 m

$x_i$ = The initial length of the spring.

= 1.25−0.60

=0.65 m

y=The collar then travels downward a distance.

= 1.80 m.

m= The mass of the collar.

=3.55 kg

$v_c$ = the velocity of the collar.

= 3.39 m/s.

g = The acceleration due to gravity.

= 9.81 m/s²

We have to calculate the magnitude of the average friction force exerted on the collar = F

Now we put the known values in the above equation, we get;

$\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c} = mgy$

Or, $\frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2} = 3.55\times 9.81\times 1.80$

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)= 8.641 N

Learn more about friction:

brainly.com/question/24338873

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Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .