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Aloiza [94]
2 years ago
7

Students are studying the two-dimensional motion of objects as they move through the air. Specifically, they are examining the b

ehavior of a sphere that is launched horizontally from a location above the floor with an initial velocity vo in the +3 direction, as shown in the figure. The students assume that the positive directions are along the sphere's initial velocity for horizontal motion and downward for vertical motion.
The horizontal displacement of the object from its starting point is x, and the vertical displacement of the object from its starting point is y. One of the students derives an equation for y in terms of xx and other quantities. After examining the equation, the student claims that y is proportional to x^2.

Required:
Derive an equation for the vertical coordinate y of the sphere as a function of x, v0, and physical constants, as appropriate.
Physics
1 answer:
kupik [55]2 years ago
7 0

Answer:

   y = - (½ g / v₀²)   x²

Explanation:

This is a projectile launch exercise where there is no acceleration on the x-axis so

        x = v₀ₓ t

        v₀ₓ = v₀ cos tea

        y = v_{oy} t - ½ g t2

        v_{oy} = v₀ sin θ

as the sphere is thrown horizontally, the angle is tea = 0º, so the initial velocity remains

          v₀ₓ = v₀

           v_{oy} = 0

we substitute in our equations

          x = v₀ t

          y = - ½ g t²

we eliminate the time from these equations, we substitute the first in the second

      y = - ½ g (x / v₀)²

      y = - (½ g / v₀²)   x²

this is the equation of a parabola

You might be interested in
A drag racer accelerates from rest at an average rate of +13.2 mls for a distance of 100. m. The driver coasts for 0.5 then uses
gtnhenbr [62]

Complete Question:

A drag racer accelerates from rest at an average rate of +13.2 m/s² for a distance of 100. m. The driver coasts for 0.5 s then uses the brakes and parachute to decelerate until the end of the track. If the total length of the track is 180 m, what minimum deceleration rate must the racer have in order to stop prior to the the end of the track?

Answer:

-31.92 m/s²

Explanation:

The drag races do a retiling uniform variated movement. There are 3 steps in the movement, first, it accelerates from rest until 100 m, second it coasts to 0.5 s, and then it decelerates. So, let's analyze each one of the steps:

Step 1

The initial velocity is v0 = 0 (because it was at rest), the acceleration is +13.2 m/s², and the distance ΔS = 100.0 m, so the final velocity, v, is:

v² = v0² + 2aΔS

v² = 2*13.2*100

v² = 2640

v = √2640

v = 51.38 m/s

Step 2

Know it's initial velocity is 51.38 m/s, it take 0.5s, and has the same acceleration, so, after 0.5 s, the velocity will be:

v = v0 + at

v = 51.38 + 13.2*0.5

v = 57.98 m/s

Thus, the distance it travels is:

v² = v0² + 2aΔS

57.98² = 51.38² + 2*13.2*ΔS

3361.6804 = 2639.9044 + 26.4ΔS

26.4ΔS = 721.776

ΔS = 27.34 m

Step 3

The initial velocity of the drag racer is 57.98 m/s, and it travels the final distance of the track: 180 - 100 - 27.34 = 52.66 m. So, when it stops, its final velocity will be 0. The minimum deceleration must be the one that it would stop at the end of the track (less than that it would cross the final track):

v² = v0² + 2aΔS

0 = 57.98² + 2a*52.66

-105.32a = 3361.6804

a = - 31.92 m/s²

4 0
2 years ago
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.240 rev/s. The magnitude
bazaltina [42]

Explanation:

Given that,

Angular velocity = 0.240 rev/s

Angular acceleration = 0.917 rev/s²

Diameter = 0.720 m

(a). We need to calculate the angular velocity after time 0.203 s

Using equation of angular motion

\omega_{f}=\omega_{i}+\alpha t

Put the value in the equation

\omega_{f}=0.240+0.917\times0.203

\omega_{f}=0.426\ rev/s

The angular velocity is 0.426 rev/s.

(b). We need to calculate the tangential speed of the blade

Using formula of  tangential speed

v= r\omega

Put the value into the formula

v = \dfrac{0.720 }{2}\times0.426\times2\pi

v=0.963\ m/s

The tangential speed of the blade is 0.963 m/s.

(c). We need to calculate the magnitude at of the tangential acceleration

Using formula of tangential acceleration

a_{t}=r\alpha

Put the value into the formula

a_{t}=0.36\times0.917\times2\pi

a_{c}=2.074\ m/s^2

The tangential acceleration is 2.074 m/s².

Hence, This is required solution.

4 0
2 years ago
23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n
PSYCHO15rus [73]

Answer:

a)  (95.4 i^ + 282.6 j^) N , b) 298.27 N  71.3º and c)   F' = 298.27 N   θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

      sin θ = Fjy / Fj

      cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

          T = 90 -32 = 58º

         Fjy = Fj sin 58

         Fjx = FJ cos 58

         Fjx = 180 cos 58 = 95.4 N

         Fjy = 180 sin 58 = 152.6 N

Andrea's force is

         Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

       Fx = Fjx

       Fx = 95.4 N

Axis y

       Fy = Fjy + Fa

       Fy = 152.6 + 130

       Fy = 282.6 N

       F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

       F² = Fx² + Fy²

       F = √ (95.4² + 282.6²)

       F = √ (88963)

       F = 298.27 N

       tan θ = Fy / Fx

       θ = tan-1 (282.6 / 95.4)

       θ = tan-1 (2,962)

       θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

       F' = 298.27 N

       θ' = 180 + 71.3

       θ = 251.4º

4 0
2 years ago
A 0.200 kg plastic ball moves with a velocity of 0.30 m/s. It collides with a second plastic ball of mass 0.100 kg, which is mov
zzz [600]

Answer:

0.22m/s

Explanation:

The total momentum of the System is conserved. Total momentum of the system before the collision is equal to the total momentum of the system after collision. The total momentum is the sum of individual momentum of all the objects in that system.

momentum of an object = mass* velocity

Total Momentum before collision = 0.2*0.3 + 0.1*0.1= 0.07 kg⋅m/s;

Total momentum after collision = 0.1*0.26 + 0.2*x = 0.07;

Solve for x.

4 0
2 years ago
B. Complete the table to show the effect of each change on each electric quantity. (1 point)
notka56 [123]

Answer:

Effect on electric force

Multiply one charge by 2

The electric force is given by F=kq1q2/r2

From the equation, the force is directly proportional to the charge.

Hence if one charge is doubled, then the electric force is doubled.

Multiply distance by 2

The electric force is given by F=kq1q2/r2

From the equation, the force is inversely proportional to the square of the distance of separation.

If the distance is doubled, F is decreased by 22. This means that the force is multiplied by 1/4.

Effect on electric potential energy

Multiply one charge by 2.

The electric potential energy is given by U=kq1q2/r

From the equation, the electric potential energy is directly proportional to the charge q.

If one charge is doubled, the electric potential energy is doubled.

Multiply distance by 2

The electric force is given by U=kq1q2/r

From the equation, the electric potential energy is inversely proportional to the distance of separation r.

If the distance is doubled, U is divided by 2. This means that the electric potential energy is multiplied by 1/2.

Effect on potential difference

Potential difference is defined as the change in electric potential energy.

Increase in the charge causes an increase in the potential difference and an increase in the distance of separation decreases the potential difference.

4 0
2 years ago
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