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podryga [215]
3 years ago
13

Now consider a different time interval: the interval between the initial kick and the moment when the ball reaches its highest p

oint. We want to find how long it takes for the ball to reach this point, and how high the ball goes. (e) What is the component of the ball’s velocity at the instant when the ball reaches its highest point (the end of this time interval)?
Physics
1 answer:
scZoUnD [109]3 years ago
3 0

Answer:A ball is kicked from a location < 9, 0, -8 > (on the ground) with initial velocity < -11, 18, -5 > m/s. The ball's speed is low enough that air resistance is negligible. What is the velocity of the ball 0.5 seconds after being kicked? (Use the Momentum Principle!) = m/s In this situation (constant force), which velocity will give the most accurate value for the location of the ball 0.5 seconds after it is kicked? The arithmetic average of the initial and final velocities. The final velocity of the ball. The initial velocity of the ball. What is the average velocity of the ball over this time interval? avg = Use the average velocity to find the location of the ball 0.5 seconds after being kicked: = m Now consider a different time interval: the interval between the initial kick and the moment when the ball reaches its highest point. We want to find how long it takes for the ball to reach this point, and how high the ball goes. What is the y-component of the ball's velocity at the instant when the ball reaches its highest point (the end of this time interval)? vyf = m/s. Fill in the missing numbers in the equation below (update form of the Momentum Principle): mvyf = mvyi + Fnet,y?t m = m + ?mg?t How long does it take for the ball to reach its highest point? ?t = s. Knowing this time, first find the y-component of the average velocity during this time interval, then use it to find the maximum height attained by the ball: ymax = m. Now take a moment to reflect on the reasoning used to solve this problem. You should be able to do a similar problem on your own, without prompting. Note that the only equations needed were the Momentum Principle and the expression for the arithmetic average velocity.

Explanation:

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If the distance between two charged particles is increased 2.69 times, the ratio of new to old electric force is _______ times t
Tasya [4]

Answer:

F = K Q1 Q2 / R^2       force between 2 charged partices

F2 / F1 = (R1 / R2)^2 = (1 / 2.69)^2 = .139

F2 = .139 F1

8 0
2 years ago
A 65.0 kg skier slides down a 37.20 slope with mu = 0.107.<br><br>What is the friction force?
zmey [24]

Answer:

54.3N

Explanation:

The normal force is perpendicular to the slope, so:

Normal Force = cos(37.2)(9.8*65).......507.39N

F(friction)=mu*F(normal)

F(friction)=(0.107)(507.39)

F(friction)=54.3N

3 0
3 years ago
A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the
marin [14]

Answer:

A)

0.395 m

B)

2.4 m/s

Explanation:

A)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

x = initial position of spring from equilibrium position = 0.21 m

v_{i} = initial speed of the cart = 2.0 ms⁻¹

A = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m

B)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

A = amplitude of the oscillation = 0.395 m

v_{o} = maximum speed at the equilibrium position

Using conservation of energy

Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}

5 0
3 years ago
Read 2 more answers
If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t
solong [7]
Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s

Answer: 1.375 s

3 0
3 years ago
An alert driver can apply the brakes fully in about 0.5 seconds. How far would the car travel if it
dybincka [34]

Answer:

The car would travel after applying brakes is, d = 14.53 m

Explanation:

Given that,

The time taken to apply brakes fully is, t = 0.5 s

The velocity of the car, v = 29.06 m/s

The distance traveled by the car in 0.5 s, d = ?

The relation between the velocity, displacement, and time is given by the formula                

                                d = v x t    m

Substituting the values in the above equation,

                                  d = 29.06 m/s x 0.5 s

                                     = 14.53 m

Therefore, the car would travel after applying brakes is, d = 14.53 m

8 0
4 years ago
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