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3 years ago

Newton discovered that gravity behaves differently in space than it does on Earth. true or false?

1 answer:

5 0

Yes. It is true. Gravity behaves differently than on earth. In-fact, as we go above the earth surface the gravity decreases. Gravity as we go above the earth surface can be calculated using following formula:

g' = g(1-2h/R)

Hence, it is true that the gravity behaves differently than on earth.

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A photon detector captures a photon with an energy of 4.29 ✕ 10−19 J. What is the wavelength, in nanometers, of the photon?

serious [3.7K]

Answer : The wavelength of photon is, $4.63\times 10^{2}nm$

Explanation : Given,

Energy of photon = $4.29\times 10^{-19}J$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength of photon = ?

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$4.29\times 10^{-19}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}$

$\lambda=4.63\times 10^{-7}m=4.63\times 10^{-7}\times 10^9nm=4.63\times 10^{2}nm$

Conversion used : $1nm=10^{-9}m$

Therefore, the wavelength of photon is, $4.63\times 10^{2}nm$

6 0

3 years ago

Which type of fusion occurs in a high-mass star near the end stages of its life cycle?

uysha [10]

The answer is "iron fusion".

In fact, initially all stars burn hydrogen through nuclear fusion. As they run out of hydrogen, they start to burn the next heavier element, which is helium. The very massive stars continue this cycle, and when they run out of helium they start to burn the heavier elements until reaching iron. This element represents the end of the chain, because nuclear fusion of iron does not release energy, but it absorbs energy. This means that the star can't produce energy anymore and eventually it collapses.

5 0

4 years ago

Read 2 more answers

For small angles, does the pendulum's period of oscillation depend on initial angular displacement from equilibrium? Explain.

aksik [14]

Answer:

No, the pendulum's period of oscillation does not depend on initial angular displacement.

Explanation:

Given that,

For small angle, the pendulum's period of oscillation depend on initial angular displacement from equilibrium.

We know that,

The time period of pendulum is defined as

$T=2\pi\sqrt{\dfrac{l}{g}}$

Where, l = length of pendulum

g = acceleration due to gravity

So, The time period of pendulum depends on the length of pendulum and acceleration due to gravity.

It does not depend on the initial angular displacement.

Hence, No, the pendulum's period of oscillation does not depend on initial angular displacement.

6 0

4 years ago

How many neutrons does element X have if its atomic number is 33 and its mass number is 80?

nikdorinn [45]

The number of neutrons is equal to the atomic mass, or mass number, minus the atomic number:
80 - 33 = 47 neutrons.

4 0

4 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

4 years ago