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3 years ago

9.

Physics

1 answer:

mote1985 [20]3 years ago

6 0

Answer:

D. location

Explanation:

The gravitational field strength of Earth is determined by the virtue of the location within the Earth's gravitational field.

That's why all objects regardless of their mass, shape, and size free fall towards the Earth with an acceleration equal to the acceleration at that location in the absence of air resistance.

According to the gravitational force between two bodies, the force experienced by one body due to the other is independent of its own mass.

The gravitational force is given by equation

F = GMm/r²

If F is the force acting on the smaller body of mass 'm', then

F = ma

Therefore, the equation becomes,

ma = GMm/r²

a = GM/r²

The value of 'a' changes with respect to the value of 'r' such that if 'r' is the radius of the Earth, then the acceleration at a height 'h' from Earth surface is given by

a = GM/(r+h)²

Here it is clear that the acceleration at any point is only the inherent property of the Earth itself.

The gravitational field strength of Earth is determined by the virtue of the location within the Earth's gravitational field.

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A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the

Bond [772]

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

$qvB = m\frac{v^2}{r}$

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

$T=\frac{2\pi r}{v}$

Re-arranging for r

$r=\frac{Tv}{2\pi}$

And substituting into the previous equation

$qvB = m \frac{Tv^3}{2\pi}$

Solving for T,

$T=\frac{2\pi q B}{m v^2}$

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

4 0

3 years ago

a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its

lilavasa [31]

Answer:

D/H =15

Explanation:

We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
At this point, we can find the value of H, applying the following kinematic equation:

$v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)$

If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

$H = \frac{v_{0} ^{2} }{2*g} (2)$

We can use the same equation, to find the value of D, as follows:

$v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)$

In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
When we replace these values in (1), we find that v₁ = -v₀.
Replacing in (3), we have:

$(4*v_{0})^{2} - (-v_{0}) ^{2} = 2* g* D\\ \\ 15*v_{0}^{2} = 2*g*D$

Solving for D:

$D = \frac{15*v_{0} ^{2} }{2*g}$

From (2) we know that H can be expressed as follows:

$H = \frac{v_{0} ^{2} }{2*g}$

⇒ D = 15 * H

$\frac{D}{H} = 15$

3 0

3 years ago

1. I’m in the 2nd column, 4th row, and I’m a metal. Who am I? ________________ 2. I’m a very lonely nonmetal. Who am I? ________

Ghella [55]

Answer:

Explanation:

I'm in 17th column , a nometal, and a solid at room temperature. What am i

6 0

3 years ago

A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two p

Anettt [7]

Answer:

A u = 0.36c B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

u ’= (u-v) / (1- uv / c²)

Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ’= 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

u’ (1- u v / c²) = u -v

u + u ’uv / c² = v - u’

u (1 + u ’v / c²) = v - u’

u = (v-u ’) / (1+ u’ v / c²)

Let's calculate

u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)

u = 0.51c / (1 + 0.4042)

u = 0.36c

We repeat the calculation for the other piece

In this case u ’= - 0.35c

We calculate

u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)

u = 1.29c / (1- 0.329)

u = 0.961c

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3 years ago

Which of the following would have the greatest force of gravitational attraction?

Mekhanik [1.2K]

Answer:

two 5kg objects 2 meters apart✨

8 0

3 years ago