Only one force acts on an object. can the object have zero acceleration? can it have zero velocity? explain

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!!

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.

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

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.

Answer:

78.4 m/s

Explanation:

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.

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

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.

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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