The resultant force acting on an object of mass 5.0kg varies with time as shown. the object is initially at rest.

Answer:

15.49m/s

Explanation:

Using one of the equations of motion;

[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2gs

Where;

v = the final velocity of the rock.

=> This is the speed at which the rock strikes the water.

u = initial velocity of the rock.

=> The rock is just dropped from that height. That means the initial velocity is zero (0).

=> u = 0

g = the acceleration due to gravity.

=> Since the rock moves downwards, the acceleration due to gravity is positive and let it have a value of 10m/[tex]s^{2}[/tex]

=> g = 10m/[tex]s^{2}[/tex]

s = the distance covered by the rock

=> s = 12m

Substituting these values into the equation above gives

[tex]v^{2}[/tex] = [tex]0^{2}[/tex] + 2(10 x 12)

[tex]v^{2}[/tex] = 240

v = [tex]\sqrt{240}[/tex]

v = 15.49m/s

Therefore the velocity(speed) with the rock strikes the water is 15.49m/s

Answer:

At 15cm to the left of the pivot

Explanation:

The weight 0.15kg(1.5N) and 3.2N are balanced on a meter stick. This weights are parallel to each other on the stick. Questions on parallel forces acting on a body are solved using the *principle of moment."

Moment is the turning effect of force about a point.

Principle of moment states that the sum of clockwise moment is equal to the sum of anticlockwise moments.

According to the question, the meter (100cm) stick is balanced with a pivot @ 18cm mark {this is the shorter end}

If the 3.2N weight is hung at xcm from the shorter end, this weight will turn about the pivot in the anticlockwise direction.

Since moment = Force × perpendicular distance from the force

Moment of the 3.2N weight = 3.2N × x = 3.2x

The weight of the stick (1.5N) will be positioned at the center of the stick i.e @ the 50cm mark which is at 32cm to the right of the pivot.

The weight of the stick will turn in the clock wise direction

The moment of the weight in the clockwise direction = 1.5×(50-18) = 1.5×32 = 48Ncm

Using the principle of moment, we will equate both moments to have;

3.2x = 48

x = 48/3.2

x = 15cm

This means the 3.2N weight will be positioned at the 15cm mark to the left if the pivot to balance the meter stick.

By definition we have to:

A system or frame of reference are those conventions used by an observer (usually standing at a point on the ground) to be able to measure the position and other physical magnitudes such as speed and acceleration of one or several objects.

The numerical value of some magnitudes can also be relative to the reference system when we refer to the relative movement. There are always mathematical relationships between the observer and the relative magnitudes.

Answer:

C. a position from which something is observed

Answer:

[tex]W = 365.7 J[/tex]

Explanation:

As we know that work done is defined as the product of force and its displacement in the direction of the force

So here we know that

applied force by Julie is same as that of weight of her suitcase

So we will have

[tex]F = mg[/tex]

[tex]F = (8\times 9.81)[/tex]

[tex]F = 78.48 N[/tex]

now we know that the total height to which the suitcase is raised is given as

[tex]H = L sin\theta[/tex]

[tex]H = 18 sin15[/tex]

[tex]H = 4.66 m[/tex]

now the work done is given as

[tex]W = F.d[/tex]

[tex]W = 78.48(4.66)[/tex]

[tex]W = 365.7 J[/tex]

Answer: B. heat

Explanation:

The nuclear reaction involves the simultaneous fission and fusion of the heavy metal nuclei such as uranium in control conditions of the nuclear reactor so as to produce energy in the form of heat. The heat or thermal energy is used to boil the water and convert it the form of steam. The steam is used to run the turbine of the generator which produces electricity.  

Answer:

Vector

Explanation:

Acceleration is defined as the rate of change of velocity. As velocity is a vector quantity, so acceleration is also a vector quantity.

If a car starts from rest and attains some velocity after some time, then car is accelerating and the firection of acceleration is same as the direction of velocity.

If a car is moving and after applying the brakes it come to rest it means the motion of car has negative acceleration which means the direction of velocity and the direction of acceleration is opposite to each other.

Using basic kinematics and trigonometry in projectile motion, we find the horizontal and vertical velocity components, time to reach the highest point, and the maximum height. We then determine the range of the projectile and the acceleration and velocity components at the highest point.

Let's solve the question step by step:

(a) Horizontal and Vertical Components of Velocity

To find the components of the shell's initial velocity, we can use trigonometry:

• Horizontal component (vx): vx = v * cos(θ) = 70.0 m/s * cos(60.0°)

• Vertical component (vy): vy = v * sin(θ) = 70.0 m/s * sin(60.0°)

(b) Time to Reach the Highest Point

The time to reach the highest point can be determined using the formula: t = vy / g, where g is the acceleration due to gravity (9.8 m/s2).

(c) Maximum Height

The maximum height is given by H = (vy2) / (2g).

(d) Horizontal Range

The horizontal range can be calculated once we have the total time of flight and the horizontal velocity: range = vx * total time of flight.

(e) Components of Acceleration and Velocity at the Highest Point

At the highest point:

• Horizontal velocity component remains unchanged: same as initial vx.

• Vertical velocity component is 0 m/s because the projectile stops rising and is about to start falling.

• Horizontal acceleration is 0 m/s2 because no forces are acting in the horizontal direction.

• Vertical acceleration remains -9.8 m/s2 (negative because it is directed downwards).

To find the ball's speed in uniform circular motion, we can use the concept of tension, gravitational force, and trigonometry. By finding the tension in the string and using the formula for speed, we can calculate the ball's speed.

To find the ball's speed, we can use the concept of circular motion. The tension in the string provides the centripetal force that keeps the ball moving in a circle. We can use the vertical component of the tension to find the gravitational force acting on the ball. The relationship between the tension, gravitational force, and the angle can be used to solve for the speed of the ball.

Using trigonometry, we can determine that the tension is equal to the gravitational force divided by the cosine of the angle. So, T = m * g / cos(26°).

Once we have the tension, we can use it to find the speed of the ball using the formula v = √(T / m), where v is the speed, T is the tension, and m is the mass of the ball. Plugging in the values, we can calculate the speed of the ball.

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Answer:

C. Mercury → Mars → Neptune → Uranus      

Explanation:

There are 8 planets in the solar system. The smallest planet is mercury and the largest planet is Jupiter. There are 4 rocky planets and 4 gaseous planets.

The arrangement of planets in increasing of their diameter is:

Mercury

Mars

Venus

Earth

Neptune

Uranus

Saturn

Jupiter

Thus, from the given options, the correct sequence is:

C. Mercury → Mars → Neptune → Uranus  

Casey can organize physics formulas by categorizing them based on topics, arranging them sequentially, using color-coded tabs, prioritizing key formulas, and creating a concise index for quick reference during the open book test.

Casey can adopt a systematic approach to organize his physics formulas for the upcoming open book test, promoting efficiency and accessibility during the examination. Firstly, he can categorize the formulas based on the topics or chapters they belong to. This helps create a structured framework that aligns with the course syllabus, making it easier for Casey to locate specific formulas relevant to different sections of the test.

Within each category, Casey may further arrange the formulas in a logical sequence, following the order of topics covered in the class. This sequential arrangement aids in maintaining a smooth flow while navigating through the formulas during the test, reducing the time spent searching for the right information. Additionally, he could use color-coded tabs or highlighters to distinguish between different formula categories, providing a visual aid for quick identification.

Considering the importance of some formulas over others, Casey might prioritize key formulas or those frequently used in class. By placing these prominently at the beginning of each category, he ensures immediate access to crucial information. Moreover, creating a concise index or summary at the front of his formula compilation can serve as a quick reference guide, aiding Casey in quickly identifying the page numbers corresponding to specific formulas.

By implementing these strategies, Casey can enhance his organization of physics formulas, facilitating a more efficient and stress-free experience during the open book test.

The question probable maybe: What organizational method could Casey employ to effectively categorize and arrange the various physics formulas for his open book test, ensuring easy access and streamlined use during the examination?

By comparing the Waxing Crescent and Waxing Gibbous lunar phases, it can be seen that the Waxing Crescent is more illuminated while the Waxing Gibbous is less illuminated. Hence option A is correct.

The second stage of the cycle of phases is the Waxing Crescent. Once a month, this Moon phase lasts for 7.38 days before transitioning into the First Quarter phase. It rises at 9 AM and sets at 9 PM. The reason this phase is known as the Waxing Crescent is because the region of the Moon's surface that is lit resembles the shape of a crescent, and waxing refers to growth. The Earth, Moon, and Sun are practically perpendicular at this phase because it is one cycle away from the First Quarter phase. This indicates that the gravitational attraction of the tides from the Sun and Moon cancels out, resulting in a smaller tidal pull. At this time, the Earth's tides are practically at neap tide.

Hence option A is correct.

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The correct answer to compare the lunar phase of the Waxing Crescent to the Waxing Gibbous is D: The Waxing Crescent is less than half illuminated and the Waxing Gibbous is more than half illuminated.

Both phases are increasing in illumination as the visible portion of the moon grows. During the Waxing Crescent, there is a growing small portion, about 1/4, on the right side of the moon that is lit. As the moon moves towards the Waxing Gibbous phase, the illuminated portion increases to more than half, approximately 3/4, on the right side of the moon.

In both waxing phases, the angle formed by pointing one arm at the Moon and one arm at the Sun demonstrates an increase in the lit portion of the moon we see. In the Waxing Crescent phase, this angle is acute, and it becomes obtuse during the Waxing Gibbous phase, indicating the moon's journey towards a full moon, when the angle is 180°, and the entire near side of the moon, as viewed from Earth, is illuminated.

All waves, regardless of their type, carry energy from one place to another without transporting matter. They do not all carry light, matter, or particles.

All waves, regardless of whether they are light waves, sound waves, electromagnetic waves, or even water waves, carry energy. Therefore, the answer to your question is a) energy.

Waves transfer energy from one place to another without transporting matter. They do this through oscillations and disturbances in a medium. For example, when you throw a stone into a pond, the waves created by the stone do not transport the water from one side of the pond to another. Instead, they transport the energy created by the stone's impact.

Not all waves carry light, matter, or particles. Light is specifically carried by electromagnetic waves, while particles and matter are not typically associated with waves in the conventional sense.

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Answer:

x is the displacement of spring

Explanation:

The Hooke's law gives the force acting on the spring when it is compressed or stretched. The mathematical expression for force is given by :

[tex]F=-kx[/tex]

Where

k is the spring constant of and it shows the stiffness of spring.

x shows the displacement of spring when it is stretched or compressed from equilibrium position and it acts in opposite direction.

Explanation of how a lever works and how to calculate its mechanical advantage when lifting a load with a given input force.

A lever is a simple machine that consists of a rigid bar pivoted at a fixed point called the fulcrum. It helps in exerting a smaller force over a longer distance to lift a heavier load over a shorter distance.

The mechanical advantage of a lever is calculated by dividing the length of the effort arm by the length of the load arm (MA = Le/Lr). It tells how many times a lever multiplies the input force to lift a load.

In the described scenario, with an input force of 500 N and a load of 2000 N, the mechanical advantage of the lever can be calculated using the formula MA = Load / Effort, which will be 4.

Answer:

a. 2.545 *10^8 J

b. 2.545 *10^8 J

c. 2.545 *10^8 J

d. 141.688 m/s

Explanation:

Data:

mass, m =  2.5*10^4 kg

Force, F = 5.00*10^5 N

distance, d = 509 m

a. Work, W = F*d = 5.00*10^5 N * 509 m = 2.545 *10^8 J

b. Change in kinetic energy, ΔKE = W = 2.545 *10^8 J

c. Change in kinetic energy = final kinetic energy - initial kinetic energy

If the train started from rest, initial kinetic energy = 0

Then, final kinetic energy, KEf = ΔKE = 2.545 *10^8 J

d. From the definition of kinetic energy we can get the final speed (vf) as follows:

KEf = (1/2)*m*vf^2

vf = √(KEf*2/m)

vf = √(2.545 *10^8*2/2.5*10^4 )

vf = 141.688 m/s

Pressure is the force per unit area. It is calculated by dividing the force applied on an object by the area over which the force is distributed.

Pressure is defined as force per unit area.

Pressure is calculated by dividing the force applied on an object by the area over which the force is distributed.

For example, if you push down on a table with a force of 50 Newtons over an area of 2 square meters, the pressure on the table is 25 Newtons per square meter.

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The correct statements are:

B. a small rock sitting on top of a big rock

As the rock is at a height with respect to ground it has potential Energy

and

C. a stretched rubber band

A stretched rubber band has elastic potential energy

The others are actually moving and hence would consist of Kinetic energy. Potential energy is stored in objects that do not move and are stationary.

Final answer:

The rate at which the area of the triangle increases when the length of one of the equal sides is 10mm is 0 mm²/s.

Explanation:

To find the rate at which the area of the triangle is increasing, we can use the formula for the area of a right triangle, which is (1/2) * base * height. Since the triangle is isosceles, the base and height are equal. Let's call the length of the equal side x. The hypotenuse is also related to x by the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

We can set up the equations:

x² + x² = c²

2x² = c²

Taking the derivative of both sides concerning time, we get:

4x * (dx/dt) = 2c * (dc/dt)

Plugging in the given values, where dx/dt = 0 (since x is constant) and dc/dt = 2mm/s, we can solve for the rate at which the area is increasing:

4(10) * (0) = 2(10.3) * (dA/dt)

0 = 20.6 * (dA/dt)

(dA/dt) = 0 mm²/s