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3 years ago

What is the gravitational force of attraction between two asteroids in space, each with a mass of 50,000 kg, separated

Physics

1 answer:

Fynjy0 [20]3 years ago

5 0

Answer:

1.16 x 10⁻⁸N

Explanation:

Given parameters:

Mass of the two asteroids = 50000kg

Distance between the asteroids = 3800m

Unknown:

Gravitational force between the two asteroids = ?

Solution:

To solve this problem, we apply the newton's law of universal gravitation:

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

G is the universal gravitation constant

r is the distance between them

So;

Fg = $\frac{6.67 x 10^{-11} x 50000 x 50000}{3800^{2} }$ = 1.16 x 10⁻⁸N

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A stargazer has an astronomical telescope with an objective whose focal length is 195 cm and an eyepiece whose focal length is 1

Colt1911 [192]

Answer:

The best angular magnification is

-375

Explanation:

Fe = Fo - L

From the question, (Fo) = 1.08 cm while (L) = 0.56cm

Thus, Fe = 1.08 - 0.56 = 0.52cm

Now, let's apply the angular magnification equation ;

M = - Di/Do

Where Di is the distance from the lens to the image and Do is the distance of the object to the lens.

In this question, Di is represented by f which is the distance from the lens to the image and it is f= 195cm

Also,Do is represented by Fe which is 0.52cm and it is the distance of the object to the lens.

Thus m = - f/fe = - 195/0.52 = - 375

6 0

4 years ago

“Blank” is a subtype of the continental climate.

andrezito [222]

D. Humid Subtropical which is like hot subtype locate in climate zone with a 45 degree north to 50 degrees north which mostly in the east to 100th meridian

8 0

4 years ago

Read 2 more answers

A piano tuner stretches a steel piano wire with a tension of 1070 N . The wire is 0.400 m long and has a mass of 4.00 g . A. Wha

pychu [463]

A. 409 Hz

The fundamental frequency of a string is given by:

$f_1=\frac{1}{2L}\sqrt{\frac{T}{m/L}}$

where

L is the length of the wire

T is the tension in the wire

m is the mass of the wire

For the piano wire in this problem,

L = 0.400 m

T = 1070 N

m = 4.00 g = 0.004 kg

So the fundamental frequency is

$f_1=\frac{1}{2(0.400)}\sqrt{\frac{1070}{(0.004)/(0.400)}}=409 Hz$

B. 24

For this part, we need to analyze the different harmonics of the piano wire. The nth-harmonic of a string is given by

$f_n = nf_1$

where $f_1$ is the fundamental frequency.

Here in this case

$f_1 = 409 Hz$

A person is capable to hear frequencies up to

$f = 1.00 \cdot 10^4 Hz$

So the highest harmonics that can be heard by a human can be found as follows:

$f=nf_1\\n= \frac{f}{f_1}=\frac{1.00\cdot 10^4}{409}=24.5 \sim 24$

8 0

4 years ago

Once out in space, the space shuttle needs to ________________ to keep going in the same direction at the same speed.

max2010maxim [7]

It is a turn off engine

8 0

3 years ago

Read 2 more answers

A cosmic ray proton moving toward the Earth at 5.00 10 m/s 7 × experiences a magnetic force of 1.70 10 N −16 × . What is the str

EastWind [94]

Answer:

Magnetic field will be equal to $B=3\times 10^{-5}T$

Explanation:

We have given velocity of proton $5\times 10^7m/sec$

Magnetic force experienced by proton $F=1.7\times 10^{-16}N$

Charge on proton $q=1.6\times 10^{-19}C$

Angle between field and velocity $\Theta =45^{\circ}$

Force in magnetic field is equal to $F=qBVsin\Theta$

So $1.7\times 10^{-16}=1.6\times 10^{-19}\times 5\times 10^7\times B\times sin45^{\circ}$

$B=3\times 10^{-5}T$

So magnetic field will be equal to $B=3\times 10^{-5}T$

7 0

3 years ago