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earnstyle [38]
3 years ago
13

the object on the right gained mass so that it had as much mass as the object on the left, how would the gravitational force bet

ween them change? Also, assume the distance between the centers of the objects stayed constant. A. It would increase. B. It would decrease. C. It would remain the same.
Physics
2 answers:
slamgirl [31]3 years ago
7 0

Answer:

a

Explanation:

kherson [118]3 years ago
6 0

The gravitational force between the objects A. It would increase.

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m_1,m_2 are the masses of the two objects

r is the separation between the objects

In this problem, we are told that one of the object (the one on the right) gains mass: this means that, for instance, the value of m_2 increases. We can see from the equation that the gravitational force is directly proportional to the masses: therefore, if one of the masses increases (while the distance between the two objects remains constant), it means that the force also increases.

Therefore, the correct answer is

A. It would increase.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light
professor190 [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

\theta_{max} =18.38^o

b

New  n_{cladding} =1.491

Explanation:

 From the question we are told that

          The refractive index of the core is  n_{core} = 1.497

         The refractive index of the cladding  is   n_{cladding} = 1.421

Generally according to Snell's law

      n_{core} * sin(90- \theta) = n_{cladding} * sin (90)

Where \theta_{max} is the largest angle a largest angle a ray will make with respect to the interface of the fiber and experience total internal reflection

      \theta_{max} = 90 - sin^{-1} [\frac{n_{cladding}}{n_{core}} ]

       \theta_{max} = 90 - sin^{-1} [\frac{1.421}{1.497}} ]

      \theta_{max} =18.38^o

Given from the question the the largest angle is  5°

Generally the refraction index of the cladding is mathematically represented as

           n_{cladding} = n_{core} * sin (90 - 5)

          n_{cladding} =1.491

       

5 0
3 years ago
If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re
jenyasd209 [6]

Answer:

The amount of work done required to stretch spring by additional 4 cm is 64 J.

Explanation:

The energy used for stretching spring is given by the relation :

E = \frac{1}{2}kx^{2}           .......(1)

Here k is spring constant and x is the displacement of spring from its equilibrium position.

For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

8 = \frac{1}{2}\timesk\times k \times(0.02)^{2}

k = 4 x 10⁴ N/m

Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).

Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

E = \frac{1}{2}\times4\times10^{4}\times (0.06)^{2}

E = 72 J

But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :

E = ( 72 - 8 ) J = 64 J

7 0
3 years ago
studies show that the ocean surface temperature is increasing which of the following is a possible consequences
katen-ka-za [31]
One possible consequence is that the warmer temperature cause the polar ice to melt even faster
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A water line with an internal radius of 5.29 x 10-3 m is connected to a shower head that has 15 holes. The speed of the water in
fiasKO [112]

Answer:

(a) 3.44 x 10^-3 m^3/s

(b) 8.4 m/s

Explanation:

area of water line, A = 5.29 x 10^-3 m

number of holes, N = 15

Speed of water in line, V = 0.651 m/s

(a) Volume flow rate is given by

V = area of water line x speed of water in water line

V = 5.29 x 10^-3 x 0.651 = 3.44 x 10^-3 m^3/s

(b) area of one hole, a = 4.13 x 10^-4 m

Let v be the velocity of water in each hole

According to the equation of continuity

A x V = a x v

5.29 x 10^-3 x 0.651 = 4.1 x 10^-4 x v

v = 8.4 m/s  

5 0
3 years ago
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