Shown below is the PV-diagram for a clockwise cycle. Determine the work done during the cycle, the heat flow during the cycle, & the change in internal energy during the cycle.

Answer:

sound A, because it has a higher frequency

took on connexus, it's correct.

Answer:

C.) The book's inertia carried it forward.

The book was at rest on the bus seat and had a tendency to stay at rest due to its inertia. When the bus stopped suddenly, the seat and the book accelerated forward due to the force applied by the bus. However, the book still had the tendency to stay at rest due to its inertia. This means that the book remained at its initial position for a brief moment while the seat and the bus moved forward. As a result, the book slid off the seat and moved forward with the same acceleration as that of the bus. So, it was the book's inertia, i.e. the resistance to change in its state of rest, that caused it to slide forward.

Answer:

C) The book's inertia carried it forward.

An object at rest tends to stay at rest, and an object in motion tends to stay in motion with the same speed and direction, unless acted upon by an unbalanced force. In this case, the book was initially at rest on the seat, but when the bus stopped suddenly, the book's inertia kept it moving forward in the same direction and speed as the bus was moving before it stopped, causing it to slide off the seat.

Explanation:

Answer:

F=w=ma                                                                                                                                                       OR by using equations of motions                                                                               vf=vi-at                    : a=vf-vi/t                                                              eq 1                         s=vit+1/2at squre                                                                                 eq 2                     2as=vf squre - vi squre                                                                        eq 3                                                                                                                                                                                  

Explanation:

where m is the mass of falling body , f is the weight is the force acting down ward , vf is the final velocity, vi is the inetial velocity , t is the time and s is the distance covered by a body.

Answer:

We can use the kinematic equations of motion to derive an equation for the time tA it takes rock A to hit the ground in terms of v0, tB, and physical constants. Since rock A is thrown vertically upward, we can use the equation:h = v0t - (1/2)gt^2where h is the initial height of the rock, v0 is the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. When rock A hits the ground, its final height is zero. So we can set h = 0 and solve for t:0 = v0tA - (1/2)gtA^2tA = (2v0)/gThis is the equation for the time it takes rock A to hit the ground. It is in terms of v0 and g, which are physical constants, and does not explicitly involve tB. However, we can use the fact that rock B hits the ground at time tB to relate tB to tA. Since rock B is thrown vertically downward, we can use the same kinematic equation with a negative value for g:h = -v0t + (1/2)gt^2When rock B hits the ground, its final height is also zero, so we can set h = 0 and solve for tB:0 = -v0tB + (1/2)gtB^2tB = (2v0)/gThis equation is identical to the equation we derived for tA. Therefore, we can conclude that both rocks hit the ground at the same time, regardless of the direction in which they were thrown.Therefore, the equation for the time tA it takes rock A to hit the ground in terms of v0, tB, and physical constants is:tA = (2v0)/gwhere v0 is the initial velocity and g is the acceleration due to gravity.

Explanation:

The time it takes for rock A to hit the ground after being thrown upwards can be calculated using the equation tA = (v0 / g) + sqrt((v0 / g)^2 + 2h0 / g). By substituting the time it takes for rock B to hit the ground (tB), we receive tA = tB + sqrt((tB^2) + 2h0 / g) with g as the acceleration due to gravity.

In Physics, the time it takes for an object to hit the ground after being thrown upwards may be calculated with the equation tA = (v0 / g) + sqrt((v0 / g)^2 + 2h0 / g), where 'g' is the acceleration due to gravity. This equation emerges from the relationship of time and displacement in free fall motion. 'v0' was the initial upwards velocity, 'h0' is the height of the cliff, and 'g' stands for gravitational acceleration which is a constant 9.8m/s².

The time that it takes rock B (which was thrown downwards) to hit the ground may be calculated from the equation, tB = sqrt((2h0) / g). Knowing that tA equals tB, we can substitute tB into the equation for tA, which results in tA = tB + sqrt((tB^2) + 2h0 / g).

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The vector whose magnitude is 36 and inclination is 60° is v = <18, 18√(3)>.

The inclination of a vector is the angle between the vector and a reference line. In this case, the reference line is the horizontal axis. Let the components be x and y. We know that the magnitude of the vector is 36, so,

magnitude = √(x² + y²) = 36

Squaring both sides of this equation, we get,

x² + y² = 1296

We also know that the inclination is 60°. The tangent of 60° is √(3), which is equal to the ratio of the vertical component to the horizontal component of the vector,

tan(60°) = y/x

y/x= √(3)

Multiplying both sides by x, we get,

y = √(3)x

Now we can substitute y in terms of x in the equation x² + y² = 1296,

x² + (√(3)x)² = 1296

Simplifying this equation, we get,

4x² = 1296

x² = 324

Taking the square root of both sides, we get,

x = +/- 18

Since the vector is making an angle of 60° with the horizontal, it must be in the first or fourth quadrant, where x is positive. Therefore, we take x = 18. Using y = √(3)x, we get,

y = sqrt(3)18

y = 18√(3)

So the vector is,

v = <18, 18√(3)>

Therefore, the vector whose magnitude is 36 and inclination is 60° is v = <18, 18√(3)>.

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