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Marat540 [252]
3 years ago
8

A friend throws a baseball horizontally. He releases it at a height of 2.0 m and it lands 21 m from his front foot, which is dir

ectly below the point at which he released the baseball. (a) How long was it in the air? (b) How fast did he throw it?
Physics
1 answer:
dangina [55]3 years ago
4 0

Answer:

<h2>a) The base baseball was in air for 0.64 seconds.</h2><h2>b) Speed of throwing is 32.89 m/s</h2>

Explanation:

a)Consider the vertical motion of baseball

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = ?

         Displacement, s = 2 m      

     Substituting

                      s = ut + 0.5 at²

                      2 = 0 x t + 0.5 x 9.81 x t²

                      t = 0.64 seconds

The base baseball was in air for 0.64 seconds.

b)Consider the horizontal motion of baseball

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = ?

        Acceleration, a = 0 m/s²  

        Time, t = 0.64 s      

         Displacement, s= 21 m

     Substituting

                      s = ut + 0.5 at²

                      21 = u x 0.64 + 0.5 x 0 x 0.64²

                      u = 32.89 m/s

      Speed of throwing is 32.89 m/s

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An igneous intrusion whose boundaries lie parallel to the layering in the surrounding country rock is considered to be:
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<span>...a concordant intrusion. In geology, "concordant" means the same as "sill" -- or, an intrusion that has gotten in between older layers of rock (or even beds of volcanic lava). An intrusion with boundaries parallel to layering in surrounding rocks suggests this, meaning it is considered to be a concordant intrusion.</span>
7 0
3 years ago
A 90 kg astronaut Travis is stranded in space at a point 12 m from his spaceship. In order to get back to his ship, Travis throw
insens350 [35]

Answer:

Explanation:

This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:

[m_av_a+m_ev_e]_b=[m_av_a+m_ev_e]_a which, in words, is

The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:

[(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a

Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).

0 = 90.0v - 2.0 and

2.0 = 90.0v so

v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.

12 = .022t and

t = 550 seconds, which is the same thing as 9.2 minutes

7 0
3 years ago
Please help!!!!!!!!!​
Elena-2011 [213]

Answer:

v = 12.86 km/h

v = 3.6 m/s

Explanation:

Given,

The distance, d = 13.5 km

The time, t = 21/20 h

                  =  1.05 h

The velocity of a body is defined as the distance traveled by the time taken.

                                     v = d / t

                                        =  13.5 km / 1.05 h

                                        = 12.86 km/h

The conversion of km/h to m/s

                                   1 km/h = 0.28 m/s

                                     12.86 km/h = 12.86 x 0.28 m/s

                                                         = 3.6 m/s

Hence, the velocity in m/s is, v = 3.6 m/s

7 0
3 years ago
A softball is thrown from the origin of an X-Y coordinate system with an initial speed of 18 m/s at an angle of 35 degrees above
Goshia [24]
Vo = 18 m/s
angle 35 degrees

1) Components of the initial velocity
 
Vox = Vo*cos(35) = 18*cos(35) m/s = 14.74 m/s
Voy = Vo* sin(35) = 18*sin(35) m/s = 10.32 m/s

2) Equations of postion:

x = Vox*t
y = Voy*t - gt^2 / 2

3) Calculations

A) t = 0.5 s, t = 1.0 st = 1.5 s, t = 2.0 s

x = 14.74 * t

t = 0.5 s => x = 14.74 m/s * 0.5s = 7.37 m

t = 1.0 s => x = 14.74 m/s * 1.0s = 14.74 m

t = 1.5s => x = 22.11 m

t = 2s => x = 29.48 m

B)

y = Voy*t - gt^2 / 2

Voy = 10.32 m/s
g = 10 m/s (approximation)

y = 10.32*t - 5t^2

t = 0.5 s=> y = 3.91m

t = 1 s => y = 5.32m

t = 1.5 s => y = 4.23m

t = 2 s => y = 0.64 m



7 0
3 years ago
How many photons per second are emitted by a monochromatic lightbulb (650 nm) that emits 45 W of power? Express your answer usin
Fudgin [204]

Answer:

The number of photons per second are 7.95\times10^{11}\ photons/s.

Explanation:

Given that,

Wavelength = 650 nm

Power = 45 W

Distance R= 17 m

Diameter = 5.0 mm

We need to calculate the number of photon per second emitted by light bulb

Using formula of energy

E=\dfrac{hc}{\lambda}

The power is

P=\dfrac{nE}{t}

\dfrac{n}{t}=\dfrac{P}{E}

Put the value of E

\dfrac{n}{t}=\dfrc{P \lambda}{hc}

Put the value into the formula

\dfrac{n}{t}=\dfrac{45\times650\times10^{-9}}{6.6\times10^{-34}\times3\times10^{8}}

\dfrac{n}{t}=1.47\times10^{20}\ photons/s

We need to calculate the surface area

Using formula of area

A=4\piR^2

A=4\pi\times17^2

We need to calculate the number of photons entering into eye

N=n\dfrac{A_{eye}}{A_{surface}}

N=1.47\times10^{20}\times\dfrac{\pi(2.5\times10^{-3})^2}{4\pi\times17^2}

N=7.95\times10^{11}\ photons/s

Hence, The number of photons per second are 7.95\times10^{11}\ photons/s.

4 0
3 years ago
Read 2 more answers
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