A cannonball fired with an initial speed of 40 m/s and a launch angle of 30 degrees from a cliff that is 25m tall.

a) What is the flight time of the cannonball?
b) What is the range of the cannonball

I have the answers i just dont know how to get there

To calculate the horizontal distance from a target at which a pilot must release a bomb, consider the time it takes for the bomb to hit the ground using projectile motion equations, then multiply this time by the airplane's horizontal speed.

The question involves calculating the horizontal distance at which a pilot must release a bomb to hit a target, considering the airplane flies horizontally with a speed of 300 m/s at an altitude of 400 m. This problem requires understanding of projectile motion, specifically the equations of motion under constant acceleration and the independence of vertical and horizontal motions. The key to solving this problem lies in determining the time it takes for the bomb to hit the ground after being released and then using this time to calculate the horizontal distance traveled.

First, to find the time (t) it takes for the bomb to fall 400 m, we use the equation of motion s = ut + 1/2 at^2, where s is the distance (400 m), u is the initial velocity (0 m/s, since the bomb is released without any initial vertical velocity), a is the acceleration due to gravity (approximated as 9.81 m/s^2), and t is the time. Solving for t gives us the time taken for the bomb to fall to the ground.

Next, since the horizontal speed of the bomb (300 m/s) remains constant (ignoring air resistance), we can calculate the horizontal distance (d) traveled by multiplying the horizontal speed by the time calculated from the first step. This gives us the horizontal distance from the release point to the target.

To summarize, by calculating the time it takes for the bomb to fall, and knowing the horizontal velocity of the airplane, we can determine the exact point at which the bomb should be released to hit the target accurately.

Final Answer:

a) The average velocity for the trip is 13.42 m/s, directed at an angle of approximately 243.4° south of west. b) The car's average acceleration during these 3.0 s is approximately 4.47 m/s², directed toward the center of the circle. c) Even though the car's speed is constant, its velocity is changing, resulting in a nonzero average acceleration.

Explanation:

a) To find the average velocity, we need to find the total displacement and divide it by the total time. The total displacement is [tex]\( \frac{3}{4} \)[/tex]of the circumference of the circle, which is [tex]\( \frac{3}{4} \times 2\pi \times 20.0 \)[/tex] m.

Thus, the total displacement is approximately 47.12 m. Dividing this by the total time of 3.0 s gives us the magnitude of the average velocity: [tex]\( \frac{47.12 \, \text{m}}{3.0 \, \text{s}} = 15.71 \, \text{m/s} \)[/tex].

To determine the direction of the average velocity, we use trigonometry. The angle [tex]\( \theta \)[/tex] can be found using the arctangent function: [tex]\( \theta = \tan^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right) = \tan^{-1}\left(\frac{20.0 \, \text{m}}{47.12 \, \text{m}}\right) \)[/tex].

Therefore, [tex]\( \theta \approx 63.4^\circ \)[/tex] south of west. Since the car's final velocity is south and its initial velocity is west, the angle of the average velocity is [tex]\( 180^\circ + 63.4^\circ = 243.4^\circ \)[/tex] south of west.

b) The average acceleration can be calculated using the change in velocity divided by the total time.

Since the car's initial and final velocities are perpendicular and the motion is circular, the acceleration is directed toward the center of the circle. The change in velocity can be found by subtracting the initial velocity vector from the final velocity vector.

The magnitude of the change in velocity is [tex]\( \sqrt{v_{fx}^2 + v_{fy}^2} \)[/tex], where [tex]\( v_{fx} \)[/tex] is the final velocity in the x-direction (south) and [tex]\( v_{fy} \)[/tex] is the final velocity in the y-direction (west).

Thus, [tex]\( v_{fx} = 0 \)[/tex] m/s and [tex]\( v_{fy} = -15.71 \)[/tex] m/s. Therefore, [tex]\( \Delta v = \sqrt{0^2 + (-15.71)^2} \, \text{m/s} \)[/tex]. Dividing this by the total time of 3.0 s gives us the magnitude of the average acceleration: [tex]\( \frac{\Delta v}{3.0 \, \text{s}} = \frac{15.71 \, \text{m/s}}{3.0 \, \text{s}} = 4.47 \, \text{m/s}^2 \)[/tex].

c) Even though the car's speed is constant, its velocity is changing because it is moving along a curved path. Velocity includes both speed and direction, so any change in direction, even if the speed remains constant, results in acceleration.

Therefore, the car experiences a nonzero average acceleration due to its change in direction while maintaining a constant speed.

Final answer:

To find the maximum load a lever can lift given its efficiency and distances from the fulcrum, calculate the mechanical advantage, then adjust for efficiency. The maximum load with 80 percent efficiency is 53.3 N, answer D.

Explanation:

The question is about calculating the maximum load that a lever can lift, given its efficiency, the effort applied, and the distances from the fulcrum on both the effort and the resistance arms. To solve this problem, we use the concept of mechanical advantage (MA) and take into account the efficiency of the lever.

The mechanical advantage (MA) of a lever is the ratio of the effort arm length to the resistance arm length. Therefore, MA = Effort Arm / Resistance Arm. In this case, MA = 2.5m / 1.5m = 1.6667.

Since the lever is at 80 percent efficiency, the actual maximum load will be less than the theoretical maximum load. If the effort is 40 N, then theoretically, the lever could lift a load of MA × Effort = 1.6667 × 40 N = 66.667 N. Considering the 80 percent efficiency, the actual maximum load that can be lifted is 0.8 × 66.667 = 53.3 N. Therefore, the correct answer is D. 53.3 N.

Final answer:

The direction in which the torpedo should be fired, as well as the time it will take to reach the battleship, requires solving problems involving relative velocity and kinematics, common in high school physics.

Explanation:

The question is related to the direction and time taken for a torpedo to hit a moving battleship which involves relative velocity and kinematics, a typical topic in high school physics. The submarine should fire the torpedo at an angle calculated using the relative speed and direction of both the torpedo and the battleship to ensure they intersect at a point. The time taken by the torpedo to reach the battleship can then be determined using the relative speeds and the initial distance between them.

An object with only one force acting can indeed have zero acceleration and zero velocity. According to Newton's first law, zero net external force on an object will maintain its state of motion, implying constant velocity (including a zero velocity) and zero acceleration.

Yes, an object with only one force acting on it can have both zero acceleration and zero velocity. Based on Newton's first law, if the net external force acting on an object is zero, the object will maintain its state of motion. This implies it can have a constant velocity, including zero velocity. Acceleration is a measure of the change in velocity, thus, if the velocity of the object remains constant, the acceleration is zero.

For instance, consider a car parked on a flat road. The car experiences forces: gravity pulling it down and the normal force (the force exerted by the surface that opposes the weight) pushing it up. These forces are balanced, making the net force zero.

Hence, the acceleration of the car is zero, and in this case, the velocity is also zero as the car is at a standstill. However, when the car is moving at a constant speed, despite any existing forces, if they are balanced and the net external force is zero, the velocity remains constant and acceleration is zero.

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Answer:

27 km/h

Explanation:

Let's take the direction in which Diego is travelling as positive direction. Therefore:

- Diego is travelling at a velocity of +7 km/h, towards the net

- The ball travels at +20 km/h towards the net, with respect to Diego's reference system

Therefore, the velocity of the ball with respect to the net reference system is equal to the sum of the two velocities:

[tex]v'=7 km/h+20 km/h=+27 km/h[/tex]

Answer:

27 km/h

Explanation: Have an amazing day and happy Easter!!

I would've thought all of the above. But if its one answer then go with budgets.

Answer: 11.025

Explanation:

Let h = the distance from the edge of the wall to the water surface (m).

Use g = 9.8 m/s² and neglect air resistance.

The initial vertical velocity of the pebble is zero.

Because the pebble hits the surface of the water after 1.5 s, therefore

h = (1/2)*(9.8 m/s²)*(1.5 s)² = 11.025 m

To find the force of gravity, identify the two masses and the distance between them, draw a free-body diagram, and apply Newton's second law of motion or Newton's universal law of gravitation.

To find the force of gravity, you need to identify the two masses involved and the distance between their centers of mass. Following this, you can draw a free-body diagram representing the forces acting on each mass.

Newton's second law of motion is then applied to each mass to determine how it will move. This is done using the equation F = ma, where F is the force, m is the mass and a is the acceleration due to gravity.

You may also use Newton's universal law of gravitation, represented by the equation F = GmM, where F is the gravitational force, G is the gravitational constant, and m and M are the masses of the two objects. Bear in mind that the force of gravity decreases with increasing distance between the objects in proportion to the inverse square of their separation.

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Splitting an atomic nucleus, also known as nuclear fission, results in the formation of two new atoms different from the original. These new atoms are often smaller, more stable, and radioactive.

When an atomic nucleus is split, the process typically results in the formation of two new atoms which are different from the original atom. This process is referred to as nuclear fission. It's important to note that nuclear fission often results in atoms that are smaller and more stable than the original, but they are not necessarily identical. These new atoms are often radioactive, meaning they emit radiation. Additionally, an enormous amount of energy is usually released during the process.

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The answer is true : )

The upward force exerted on the passenger is 836 N.

The given parameters;

The upward force on the object can be determined by applying Newton's second law of motion.

Since the elevator is ascending upwards, the force on the elevator will become greater and the passenger feels a heavier weight.

∑F = ma + mg

F = m(a + g)

F = 76(1.2 + 9.8)

F = 836 N.

Thus, the upward force exerted on the passenger is 836 N.

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Answer:

78.4 m/s

Explanation:

Answer:

B) 85

Explanation:

In the nuclear fission of uranium-235, a nucleus of uranium-235 absorbs a slow-moving neutron and then decays into a nucleus of Kripton-92, a nucleus of Barium-141 (which is the missing nucleus in the figure) and three more neutrons.

Knowing that:

- Uranium has atomic number 92, so it has 92 protons

- Kripton has atomic number 36, so it has 36 protons

- Barium has atomic number 56, so it has 56 protons

We can calculate the number of neutrons in the nucleus of barium, since we know that the total number of neutrons must be conserved. We have:

- On the left side of the reaction: (235-92) neutrons of the uranium + 1 neutron absorbed = 144 neutrons

- On the right side of the reaction: (92-36) neutrons of the krypton + X neutrons of the barium + 3 neutrons = 144

So we get:

56 + X + 3 = 144

X = 85

So, the nucleus of barium has 85 neutrons.

Light rays entering a pinhole camera form an inverted image on its back wall; the idea is inverted because light travels in straight lines and doesn’t reorient upon entering the pinhole.

Light rays that enter the hole of a pinhole camera form an inverted image on the back wall of the camera. This occurs as the light rays from an object diverge, and only a tiny portion of these rays enter the pinhole, causing them to form an image. Notably, this image is flipped upside down - or inverted - because light travels in straight lines, and the rays do not reorient themselves after entering the pinhole. An important aspect to understand is that these rays do not cross each other upon entering the pinhole, thereby ensuring the formation of a clear and inverted picture on the back wall of a pinhole camera. This is fundamentally akin to how light enters the human eye to form an image on the retina, where a similar effect is also observed, resulting in the need for the brain to flip the image right-side up for us to perceive our surroundings correctly.

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