1.13g plastic stopper is attached to a 0.93-m string. The stopper is swung in a horizontal circle, making one revolution in 1.18s(period T). Find the tension force(equivalent to the centripetal force)exerted by the string on the stopper? (V=2(pie)r/T)?
2. A football is thrown at 30 m/s(speed of release Vi)is 30 m/s, and its angle of release is 40 degree with respect to the horizontal. What is its horizontal velocity Vix? And what is its vertical velocity Viy? (Using trigonometry)
3. In a long jumper's motion, the highest point of the trajectory Y is 1.8m, the horizontal velocity at this point Vx is 7.2m/s. What is the full range of this long jump motion?(notice that the trajectory for this motion is symmetrial)
4. A car traveling at 30 m/s lost control and fell off a 50-m cliff by a projectile trajectory into a flat valley. Where should the car land on the flat bottom of the valley?
(( Note I was absent but my teacher couldn't explain anything to me so help please?))
When you post serveral questions on the same post, without any work or indication of what you need, your chances go down.
a. Tension=mass*velocity^2/r
and your teacher is just too easy with this hint: (V=2(pie)r/T)
b. vix=speed*cosTheta
vertical speed=speed*sinTheta
c.time to fall:
h=1/2 g t^2
t= sqrt (2*1.8/9.8)
range=2t*7.2 m
d. time to fall:
50=1/2 g t^2
t= sqrt(100/9.8)
distance= 30m/s * time
So for the first one I got 0.78.
The second 23 horizontal and 20 vertical.
The third I got 8.64
And fourth I got 96
??
1. To find the tension force exerted by the string on the stopper, you can use the equation:
T = m * (v^2 / r)
where T is the tension force, m is the mass of the stopper, v is the velocity of the stopper, and r is the radius of the circular path.
In this case, the mass of the stopper is given as 1.13g, which is equivalent to 0.00113kg. The period of revolution (T) is given as 1.18s. To find the velocity (v), you can use the equation:
v = 2 * π * r / T
where π is a constant approximately equal to 3.14159.
Plugging in the given values:
v = 2 * π * r / T
v = 2 * 3.14159 * 0.93m / 1.18s
v ≈ 4.99 m/s
Now you can plug the values of m and v into the tension force equation:
T = m * (v^2 / r)
T = 0.00113kg * (4.99 m/s)^2 / 0.93m
T ≈ 0.00608 N
Therefore, the tension force exerted by the string on the stopper is approximately 0.00608 Newtons.
2. To find the horizontal velocity (Vix) and vertical velocity (Viy) of the football, you can use the following equations:
Vix = Vi * cos(angle)
Viy = Vi * sin(angle)
where Vi is the initial velocity and angle is the angle of release with respect to the horizontal.
In this case, the initial velocity (Vi) is given as 30 m/s and the angle of release is given as 40 degrees.
Plugging in the given values:
Vix = 30 m/s * cos(40 degrees)
Vix ≈ 22.99 m/s
Viy = 30 m/s * sin(40 degrees)
Viy ≈ 19.28 m/s
Therefore, the horizontal velocity (Vix) is approximately 22.99 m/s and the vertical velocity (Viy) is approximately 19.28 m/s.
3. Given that the highest point of the long jumper's trajectory (Y) is 1.8m and the horizontal velocity at this point (Vx) is 7.2m/s, you can find the full range of the long jump motion.
The trajectory of the long jump motion is symmetric, so the horizontal velocity remains constant throughout the motion. Therefore, the range is given by the equation:
Range = 2 * Vx * t
where Vx is the horizontal velocity and t is the time it takes for the jumper to reach the highest point.
To find the time (t), you can use the equation:
t = Y / (Vy)
where Y is the highest point of the trajectory and Vy is the vertical velocity at that point.
In this case, Y is given as 1.8m and Vx is given as 7.2m/s. From the previous calculations, we know that Vy = 0 at the highest point.
Plugging in the given values:
t = 1.8m / (0m/s)
t = Infinity
The time is undefined because there is no vertical component of velocity at the highest point. This means that the long jumper will stay at the highest point for an infinite amount of time before falling back down.
Therefore, the full range of the long jump motion is also undefined.
4. To find where the car should land on the flat bottom of the valley after falling off a 50m cliff, you can use projectile motion equations.
Since the car is falling under the influence of gravity, its initial vertical velocity (Vi) is 0 m/s, and the only force acting on it is gravity.
Using the equation:
s = Vi * t + (1/2) * g * t^2
where s is the vertical distance fallen, Vi is the initial vertical velocity, t is the time taken to fall, and g is the acceleration due to gravity (-9.8 m/s^2).
In this case, the vertical distance fallen is 50m. Plugging in the given values:
50m = 0m/s * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying the equation:
-4.9t^2 = 50m
Solving for t:
t^2 = 50m / -4.9
t^2 ≈ 10.2s^2
Taking the square root of both sides:
t ≈ √10.2s^2
t ≈ 3.2s
Now that we know the time taken to fall, we can find the horizontal distance (range) that the car will travel during this time.
Using the equation:
Range = horizontal velocity * time
Given that the car's initial horizontal velocity is not provided, we cannot determine the exact horizontal distance. However, assuming the car maintains a constant horizontal velocity, you can use an average velocity to estimate the landing point.
For example, if the car maintains an average horizontal velocity of 30 m/s throughout its fall, then:
Range = 30 m/s * 3.2s
Range ≈ 96m
Therefore, if the car has an average horizontal velocity of 30 m/s, it should land approximately 96 meters from the cliff base on the flat bottom of the valley.
Sure, I'd be happy to help you with these physics problems!
1. To find the tension force exerted by the string on the plastic stopper, we need to use the equation for centripetal force, which is given by F = mv²/r, where m is the mass of the stopper, v is the linear velocity, and r is the radius of the circular path. In this case, m = 1.13 g = 0.00113 kg. The linear velocity can be calculated using v = 2πr/T, where T is the period of one revolution. Given T = 1.18 s and r = 0.93 m, we can substitute these values into the equation and find v. Finally, plug the values of m, v, and r into the equation F = mv²/r to find the tension force.
2. To find the horizontal and vertical velocities of a thrown object, we can use the equations for projectile motion. The horizontal velocity (Vix) remains constant throughout the motion and is given by Vix = Vi * cos(θ), where Vi is the initial velocity and θ is the angle of release. In this case, Vi = 30 m/s and θ = 40 degrees. Plug these values into the equation to calculate Vix. The vertical velocity (Viy) changes due to the effect of gravity. Viy can be calculated using Viy = Vi * sin(θ). Again, substitute the given values into the equation to find Viy.
3. To find the full range of a long jump motion, we need to calculate the total horizontal distance covered by the jumper. The range can be determined by multiplying the horizontal velocity (Vx) at the highest point of the trajectory by the total time taken to reach the highest point and back down to the same level. Since the trajectory is symmetrical, the total time is twice the time it takes to reach the highest point. Given Vx = 7.2 m/s, the time to reach the highest point can be found by dividing the vertical displacement by Viy, and then multiplying it by 2 (since the motion is symmetrical). Once you have the time, multiply it by Vx to find the range.
4. To find where the car lands on the flat bottom of the valley, we can analyze the projectile motion of the car after it falls off the cliff. The horizontal position of the car when it lands can be determined by multiplying the horizontal velocity (Vx) of the car by the time it takes to reach the height of the cliff. The time can be found by dividing the height of the cliff (50 m) by the vertical velocity (Viy) of the car. Use the equation Vf² = Vi² + 2ad, where Vi is the initial vertical velocity of the car (which can be calculated using the equation Vi = sqrt(2gh), where g is the acceleration due to gravity and h is the height of the cliff), a is the acceleration (which is equal to -g), Vf is the final vertical velocity (which is zero when the car lands), and d is the vertical displacement (which is equal to -h since it is a downward motion). Once you have the time, multiply it by Vx to find the horizontal position where the car lands.
I hope these explanations help you understand how to solve these physics problems. Let me know if you have any further questions!