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Liula [17]
3 years ago
14

Two spherical objects have masses of 3.1 x 10^5 kg and 6.5 x 10^3 kg. The gravitational attraction between them is 65 N. How far

apart are their centers? (answer to 2 digits)
Physics
1 answer:
nata0808 [166]3 years ago
3 0

Answer:

4.55 x 10⁹m

Explanation:

Given parameters:

Mass of object 1  = 3.1 x 10⁵kg

Mass of object 2 = 6.5 x 10³kg

Gravitational force  = 65N

Unknown:

Distance between them  = ?

Solution:

To solve this problem, we use the expression below from the universal gravitational law;

    Fg  =    \frac{G mass 1 x mass 2}{distance ^{2} }  

   G = 6.67 x 10⁻¹¹

        65  = \frac{6.67 x 10^{11} x 3.1 x 10^{5} x 6.5 x 10^{3}   }{distance^{2} }    

   Distance  = 4.55 x 10⁹m

         

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Whenever a musician plays a guitar, they pluck one of the guitar strings to produce a standing wave in the string. Then pinch do
UkoKoshka [18]

Answer:

It's due to the distance from either ends of strings origin...

Explanation:

As we know that waves behave moving in a flow from one side to another side and this gives a prospective of motion. Suppose a wave is pinched from the near one end of a guitar then due to the distortion created by the point of tie of strings the wave super imposes and moves with a velocity v and produces a wave frequency f. as we the pinching go down to the center the wave stabilizes itself to a stationary origin right at the center and the frequency then changes accordingly as moving down on the string.

7 0
3 years ago
A lab technician uses laser light with a wavelength of 650 nmnm to test a diffraction grating. When the grating is 42.0 cmcm fro
Bingel [31]

Answer:

221 lines per millimetre

Explanation:

We know that for a diffraction grating, dsinθ =mλ where d = spacing between grating, θ = angle to maximum, m = order of maximum and λ = wavelength of light.

Since the grating is 42.0 cm from the screen and its first order maximum (m = 1) is at 6.09 cm from the center of the pattern,

tanθ = 6.09 cm/42.0 cm = 0.145

From trig ratios, cot²θ + 1 = cosec²θ

cosecθ = √((1/tanθ)² + 1) = √((1/0.145)² + 1) = √48.562 = 6.969

sinθ = 1/cosecθ = 1/6.969 = 0.1435

Also, sinθ = mλ/d at the first-order maximum, m = 1. So

sinθ = (1)λ/d = λ/d

Equating both expressions we have  

0.1435 = λ/d

d = λ/0.1435

Now, λ = 650 nm = 650 × 10⁻⁹ m

d = 650 × 10⁻⁹ m/0.1435

d = 4529.62 × 10⁻⁹ m per line

d = 4.52962 × 10⁻⁶ m per line

d = 0.00452962 × 10⁻³ m per line

d = 0.00452962 mm per line

Since d = width of grating/number of lines of grating

Then number of lines per millimetre = 1/grating spacing

= 1/0.00452962

= 220.77 lines per millimetre

≅ 221 lines per millimetre since we can only have a whole number of lines.

6 0
3 years ago
1.) A ski jump is angled at 40° and the skier is launched at 25 m/s. A.) What was her highest
prohojiy [21]

A) 13.2 m

The motion of the skier is a projectile motion, which consists of two independent motions:

- a horizontal motion with constant speed

- a vertical motion with constant acceleration g=-9.8 m/s^2 (acceleration of gravity) downward

To find the maximum height of her trajectory, we are only concerned with her vertical motion.

The initial vertical velocity upward is

u_y = u_0 sin \theta = (25) sin 40^{\circ} =16.1 m/s

then we can use the following SUVAT equation:

v_y^2 - u_y^2 = 2ah

where

v_y=0 is the final vertical velocity, which is zero at the maximum height

u_y = 16.1 m/s is the initial vertical velocity

a=g=-9.8 m/s^2

h is the maximum height

Solving for h,

h=\frac{v_y^2-u_y^2}{2g}=\frac{-(16.1)^2}{2(-9.8)}=13.2 m

B) 1.64 s

The time needed to reach the highest point can be found by analyzing again the vertical motion only. In fact, we can use the  following SUVAT equation:

v_y = u_y +at

where

u_y = 16.1 m/s

a=g=-9.8 m/s^2

At the maximum height, the vertical velocity is zero:

v_y=0

So we can solve the equation to find the corresponding time:

t=\frac{v_y-u_y}{a}=\frac{0-16.1}{-9.8}=1.64 s

7 0
4 years ago
What is the correct answer?
katen-ka-za [31]
Red line -1 because it is just starting out and climbing speed so his energy is at lowest hopefully this helps
4 0
3 years ago
A charged particle accelerated to a velocity v enters the chamber of a mass spectrometer. The particle's velocity is perpendicul
gladu [14]

Answer:

Circle

Explanation:

When a charged particle is in motion in a region with magnetic field, the particle experiences a force whose magnitude is given by

F=qvB sin \theta

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

\theta is the angle between the directions of v and B

In this problem, the velocity of the particle is perpendicular to the magnetic field, so

\theta=90^{\circ}

and the formula reduces to

F=qvB

Also, the direction of this force is perpendicular to the direction of motion of the particle. This means that as the charge moves in the region of the magnetic field, the force acting on it acts as a centripetal force: therefore, the particle will start moving by unifom circular motion, with constant speed (because the magnetic force does no work on the particle, since it is perpendicular to the direction of motion).

So, the path of the particle will be a circle.

4 0
4 years ago
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