Initilal velocity of projectile at an angle of 22.5° is 10m/s. Then what is the magnitude of range of projectile?

The hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would pro

Gnesinka [82]

The hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would produce is apparently as a result of <u>Eddington luminosity</u>

<h3>What are stars?</h3>

Stars are a fixed luminous point in the sky which is a large and remote incandescent body

So therefore, the hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would produce is apparently as a result of Eddington luminosity

Learn more about stars:

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4 0

2 years ago

5. A 5.0 kg object accelerates uniformly from rest for 5.0 s and reaches a final velocity of 20.0 m/s. At 3.0 s, what is the obj

irina [24]

Answer:

jack JACKKK

Explanation:

OK YAHHHHH SO DO THIS

3 0

3 years ago

A simple pendulum consists of a mass m (the "bob") hanging on the end of a thin string of length l and negligible mass. The bob

svetlana [45]

Answer:

The pendulum frequency is (c) the same, or very close to it

Explanation:

The simple pendulum corresponds to a simple harmonic movement, to reach this approximation in the expression of the force the sine of the angle (θ) approaches an angle value, this is only true for small angles, generally less than 15º

Sine (15th) = 0.2588

The angle in radians is 15º π / 180º = 0.26180.2588 / 0.2618

The difference between these two values is less than 1.2%

for smaller angle the difference is reduced more

Therefore, the period for both the 5º and 10º angles is almost the same

5 0

3 years ago

What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$