A very small round ball is located near a large solid sphere of uniform density. The force that the large sphere exerts on the b

Answer:

The amount of heat that enters the gas throughout this two-step process totals 120 cal.

Explanation:

Given that,

Moles present = 3

Heat capacity at volume held constant = 4.9 cal/mol.K

Heat capacity at pressure held constant = 6.9 cal/mol.K

Starting temperature = 300 K

Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

Substituting the values into the equation

Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

Insert the values into the formula

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.

Here's the procedure explained: Assume F represents the portion of the rope that is extending over the table. In this scenario, the frictional force that holds the rope on the table can be calculated using the formula: Ff = u*(1-f)*m*g. Additionally, it is important to determine the gravitational force that attempts to pull the rope off the table, Fg, calculated through: Fg = f*m*g. You then need to set these two equations equal to each other and resolve for f: f*m*g = u*(1-f)*m*g leads to f = u*(1-f) = u - uf. Simplifying gives f + uf = u, which results in f = u/(1+u) representing the fraction of the rope. This will lead you to the final answer.

The answer is letter d, 8.3.

 

Here’s a solution for the given problem:

 

We have:

B = 6i - 8j 

Let A be unknown; we'll denote A as = mi + nj 

The resultant A+B lies along the x-axis (which implies A+B = Ki + 0j, where K is yet to be determined,

 

and we also know the magnitude of A+B is equivalent to the magnitude of A,

 

therefore, mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2. 

Using vector addition, A+B becomes (m+6)i + (n-8)j.

 

Since we know A+B = Ki + 0j, we can establish that: 

m + 6 = K 

n - 8 = 0, which gives n=8. 

Thus, K^2=m^2+n^2 means (m+6)^2 = m^2 +8^2 

= m^2 + 12m + 36 = m^2 + 64 

which gives us 12m = 28 

m = 2.33333... 

Consequently, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333.

Answer:

1.5 × 10³⁶ light-years

Explanation:

A particular square area in interstellar space measures roughly 2.4 × 10⁷² (light-years)². To find the area of a square, the following formula is utilized:

A = l²

where,

A represents the area of the square

l denotes the length of one side of the square

Thus, l = √A = √2.4 × 10⁷² (light-years)² = 1.5 × 10³⁶ light-years

To tackle this issue, we will utilize concepts related to gravity based on Newtonian definitions. To find this value, we'll apply linear motion kinematic equations to determine the required time. Our parameters include:

Comet mass

Radius

The rock is released from a height 'h' of 1 m above the surface.

The relationship for gravity's acceleration concerning a body with mass 'm' and radius 'r' is described by:

Where G represents the gravitational constant and M denotes the mass of the planet.

Now, let’s compute the time value.

Ultimately, the time for the rock to hit the surface is t = 87.58s.