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How much wok must be done on a 6000 kg stone car to increase its speed from 5 miles per second to 10 miles per second?
To increase the speed of the stone car from 5 miles per second to 10 miles per second, you would need to do work equal to the change in kinetic energy. This can be calculated using the formula: ( \text{work} = \frac{1}{2} \times \text{mass} \times \left( \text{final velocity}^2 - \text{initial velocity}^2 \right) ). Plugging in the values for mass (6000 kg), initial velocity (5 miles per second), and final velocity (10 miles per second) will give you the amount of work required.
What is the effect of changing the initial position on position time and velocity time graph?
Changing the initial position on a position vs time graph has no effect on the velocity vs time graph. Velocity is the derivative of position. This means velocity only depends on the rate of change (slope) of position. Changing the initial position of an object has no effect on the slope. Mathematically, this is equivalent to adding a constant to a function. Since the derivative of a constant is always 0, a change in initial position has no impact on the derivative. Here is an example. Say we have the position functions x(t)= 4+9t and y(t)= 27+9t. then the velocity function of x would be x'(t)=v(t)= 9 And the velocity function of y would be Y'(t)=v(t)= 9
Why does a ball you toss into the air not fall right away to the floor?
The ball is thrown with an initial velocity, and gravity slows it down as it rises. At its peak, the ball's velocity is zero before it begins to fall back to the ground. This is due to the balance between the initial force and gravity acting on the ball.
How can you find initial velocity when the displacement is given but time is not?
Unfortunately, the question is so vague it can't be answered specifically. But I will use my intuition and say that you are a high-school student studying algebra-based physics and are now up to the chapter or unit on rectilinear motion. If so, read on. If not, well, then perhaps you could use the discussion page to add more info (and maybe rephrase the question). The general formula for straight-line motion is a quadratic equation. Displacement (or distance) is expressed as a function of time. In other words, dispacement is the dependent variable and time is the independent variable. But displacement is also dependent upon the values of initial displacement, initial velocity, and acceleration, which are all coefficients of the general displacement formula. Here is the formula: d = d0 + v0t + (1/2)at2, where d is the displacement, d0 is the initial displacement (in other words, the displacement at t = 0), v0 is the initial velocity (velocity at t = 0), and a is the acceleration. So, using the formula, you can solve for distance traveled (displacement) if you know the values of all those parameters to the right of the equal sign. But what if you don't know the value of t? Well, in that case, you had better know the values of all the other parameters, including the d to the left of the equal sign. If you know the initial distance, the total distance traveled, the initial velocity, and the acceleration, you can solve for t. Usually, we set up our frame of reference so that d0 = 0. Frequently, v0 = 0, also. (In other words, the object has no initial displacement and no initial velocity.) If you know the distance traveled, d, and the acceleration, a, then you can solve for t using the simplified formula d = (1/2)at2. Solving for t, you get t = SQRT(2d/a).Since you now know t and d, you can calculate the object's average velocity using the formula, va = d/t. Since the object started at rest (it had zero initial velocity), its final velocity, vf, is 2va. You might be able to use the equation of motion v2 = u2 + 2ad, where v is the final velocity, u is the initial velocity, a is the acceleration & d is the distance covered. Quite often the initial velocity is zero, so the equation simply becomes v2 = 2ad. So the final velocity v =SQRT(2ad).
If a ball is rolling in a straight line and you push it to the right it's velocity Will?
If you push a ball rolling in a straight line to the right, its velocity will change in the direction you pushed it, affecting its speed and direction of motion. The velocity will have a component in the direction you pushed it, as well as any remaining component from its original motion.
How do you calculate velocity?
given distance- 200km time-5 hours speed- distance/time 200/5 40km/hr Divide distance by time for speed Divide speed by distance for time Multiply speed and time for distance Those are three ways to make sure you have it right Average Velocity = (change in position) / (elapsed time) Instantaneous Velocity = [limit as elapsed time approaches 0] (change in position) / (elapsed time) Velocity is measured in m/s+ direction
How much wok must be done on a 6000 kg stone car to increase its speed from 5 miles per second to 10 miles per second?
To increase the speed of the stone car from 5 miles per second to 10 miles per second, you would need to do work equal to the change in kinetic energy. This can be calculated using the formula: ( \text{work} = \frac{1}{2} \times \text{mass} \times \left( \text{final velocity}^2 - \text{initial velocity}^2 \right) ). Plugging in the values for mass (6000 kg), initial velocity (5 miles per second), and final velocity (10 miles per second) will give you the amount of work required.
If a ball rolls off the edge of a table two meters above the floor and with an initial velocity of 20 meters per second what is the ball's acceleration and velocity just before it hits the ground?
The horizontal velocity has no bearing on the time it takes for the ball to fall to the floor and, ignoring the effects of air resistance, will not change throughout the ball's fall, so you know Vx. The vertical velocity right before impact is easily calculated using the standard formula: d - d0 = V0t + [1/2]at2. For this problem, let's assume the floor represents zero height, so the initial height, d0, is 2. Further, substitute -g for a and assume an initial vertical velocity of zero, which changes our equation to 0 - 2 = 0t - [1/2]gt2. Now, solve for t. That gives you the time it takes for the ball to hit the floor. If you divide the distance traveled by that time, you know the average vertical velocity of the ball. Double that, and you have the final vertical velocity! (Do you know why?) Now do the vector addition of the vertical velocity and the horizontal velocity. Remember, the vertical velocity is negative!
What is the effect of changing the initial position on position time and velocity time graph?
Changing the initial position on a position vs time graph has no effect on the velocity vs time graph. Velocity is the derivative of position. This means velocity only depends on the rate of change (slope) of position. Changing the initial position of an object has no effect on the slope. Mathematically, this is equivalent to adding a constant to a function. Since the derivative of a constant is always 0, a change in initial position has no impact on the derivative. Here is an example. Say we have the position functions x(t)= 4+9t and y(t)= 27+9t. then the velocity function of x would be x'(t)=v(t)= 9 And the velocity function of y would be Y'(t)=v(t)= 9
If a batter hits a baseball upward into right field does its velocity change?
yes
Why does a ball you toss into the air not fall right away to the floor?
The ball is thrown with an initial velocity, and gravity slows it down as it rises. At its peak, the ball's velocity is zero before it begins to fall back to the ground. This is due to the balance between the initial force and gravity acting on the ball.
How can you find initial velocity when the displacement is given but time is not?
Unfortunately, the question is so vague it can't be answered specifically. But I will use my intuition and say that you are a high-school student studying algebra-based physics and are now up to the chapter or unit on rectilinear motion. If so, read on. If not, well, then perhaps you could use the discussion page to add more info (and maybe rephrase the question). The general formula for straight-line motion is a quadratic equation. Displacement (or distance) is expressed as a function of time. In other words, dispacement is the dependent variable and time is the independent variable. But displacement is also dependent upon the values of initial displacement, initial velocity, and acceleration, which are all coefficients of the general displacement formula. Here is the formula: d = d0 + v0t + (1/2)at2, where d is the displacement, d0 is the initial displacement (in other words, the displacement at t = 0), v0 is the initial velocity (velocity at t = 0), and a is the acceleration. So, using the formula, you can solve for distance traveled (displacement) if you know the values of all those parameters to the right of the equal sign. But what if you don't know the value of t? Well, in that case, you had better know the values of all the other parameters, including the d to the left of the equal sign. If you know the initial distance, the total distance traveled, the initial velocity, and the acceleration, you can solve for t. Usually, we set up our frame of reference so that d0 = 0. Frequently, v0 = 0, also. (In other words, the object has no initial displacement and no initial velocity.) If you know the distance traveled, d, and the acceleration, a, then you can solve for t using the simplified formula d = (1/2)at2. Solving for t, you get t = SQRT(2d/a).Since you now know t and d, you can calculate the object's average velocity using the formula, va = d/t. Since the object started at rest (it had zero initial velocity), its final velocity, vf, is 2va. You might be able to use the equation of motion v2 = u2 + 2ad, where v is the final velocity, u is the initial velocity, a is the acceleration & d is the distance covered. Quite often the initial velocity is zero, so the equation simply becomes v2 = 2ad. So the final velocity v =SQRT(2ad).
Are rates of change always constant?
If I understand correctly your question, the answer is definitely no.Think about the typical physical representation of the concept of "rate of change": the velocity is the rate of change of position, right? And there's no difficulty in imagining a non-constant velocity, as when you accelerate or decelerate..
If a ball is rolling in a straight line and you push it to the right it's velocity Will?
If you push a ball rolling in a straight line to the right, its velocity will change in the direction you pushed it, affecting its speed and direction of motion. The velocity will have a component in the direction you pushed it, as well as any remaining component from its original motion.
What would happen to an object's velocity if no work was being done on the object?
The velocity might still change, in the case of a force applied at a right angle to the movement. In this case, since the object's direction changes, its velocity changes.
What happens when velocity and acceleration are not in the same direction?
When velocity and acceleration are not in the same direction, the object's speed may be changing. If the acceleration and velocity are in opposite directions, the object will slow down. If they are at right angles to each other, the object will change direction without changing speed.
What is the velocity of a 50 gram egg dropped 3 meters off the ground right before it hits the ground?
Ignoring the effects of air resistance, the velocity of any object dropped near the surface of the Earth is 29.42 meters (96.53 feet) per second, directed downward, after it has fallen 3 meters. If you're working with situations and asking questions like this one, then you're supposed to know by now that the mass or weight of the object makes no difference. If you can eliminate air resistance, then the answer is the same for a feather, a bowling ball, and a battleship.
