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Vinil7 [7]
3 years ago
10

A baseball is given an initial velocity with magnitude v at the angle beta above the surface of an incline which in turn incline

d at angle teta above horizontal calculate the distance measured along incline from the launch point to where the baseball strike the incline
​

Physics
1 answer:
lakkis [162]3 years ago
8 0

Explanation:

The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 ⁡ θ i 2 g .

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A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
3 years ago
A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi
vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

V=\frac {C_2V_2-C_1V_1}{C_1+C_2}

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}

Therefore

V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437

Charge, Q is given by CV hence for the first capacitor charge will be Q_1=C_1V

Here, Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C

8 0
3 years ago
1. If an object of mass m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, th
Archy [21]

Answer:

1a). 4 times. 1b) 4. 1c) 1/4.

2a) 5 times. 1b) 5 1c) 1/5

3a) 7/3 times 3b) 7/3 3c) 3/7

4a) 8/5 times 4b) 8/5 4c) 5/8

Explanation:

1) Assuming no external forces acting during the collision, total momentum must be conserved, so the following general expression applies:

m₁*v₁₀ + m₂*v₂₀ = m₁*v₁f  + m₂*v₂f (1)

If we assume that the collision is perfectly inelastic, this means that both masses stick together after the collision, so v₁f = v₂f.

If m₂ is initially at rest, ⇒ v₂₀ = 0.

Replacing in (1) we get the expression of vf as a function of v₁₀, as follows:

vf = v₁₀*(m₁/(m₁+m₂)

So, for the four cases we have the following:

1) initial mass = m

  final mass = m+3m = 4 m

⇒final mass / initial mass = 4

vf = v₀* (m/4m) = v₀/4  ⇒v₀/vf = 4

So, the velocity of the system will decrease by a factor of 4. The new velocity will be vf= v₀/4.

2) Applying the same considerations, we get:

2a)  final mass / initial mass = 5

2b) vf = v₀* (m/5m) = v₀/5  ⇒v₀/vf = 5

2c) vf = v₀/5

3) Applying the same considerations, we get:

3a)  final mass / initial mass = 7/3

3b) vf = v₀* (3m/7m) =3/7* v₀  ⇒v₀/vf = 7/3

3c) vf = 3/7*v₀

4) Applying the same considerations, we get:

4a)  final mass / initial mass = 8/5

4b) vf = v₀* (5m/8m) = 5/8*v₀ ⇒v₀/vf = 8/5

4c) vf = 5/8*v₀

3 0
3 years ago
Inside most ball-point pens is a small spring that compresses as the pen is pressed against the paper. If a force of 0.1 N compr
AnnZ [28]

Answer:

20 N/m

Explanation:

From the question,

The ball-point pen obays hook's law.

From hook's law,

F = ke............................ Equation 1

Where F = Force, k = spring constant, e = compression.

Make k the subject of the equation

k = F/e........................ Equation 2

Given: F = 0.1 N, e = 0.005 m.

Substitute these values into equation 2

k = 0.1/0.005

k = 20 N/m.

Hence the spring constant of the tiny spring is 20 N/m

8 0
3 years ago
A wave travels at a frequency of 387 Hz. What is<br>the period of the wave?<br>​
Ulleksa [173]

Answer:

¹/₃₈₇ second

Explanation:

<em>The period of a wave is the reciprocal of its frequency.</em>

So, simply, the frequency is ¹/₃₈₇ second(s), as that is the reciprocal of the frequency, 387 Hz.

5 0
3 years ago
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