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goldenfox [79]
3 years ago
5

A baseball is batted. It's a long fly ball. 4 seconds later the ball reaches the outfield 100 meters away and returns to the hei

ght from which it left the bat. Determine the horizontal component of the ball's initial velocity in magnitude, in m/s,
Physics
1 answer:
Vitek1552 [10]3 years ago
8 0

Answer:

25 m/s

Explanation:

First we should define the variables

T=4

Dx = 100

ay=-9.8

ax=0

We can use formula 1 from the BIG 5

x=(v+v0)t/2

By plugging in our variables we can get 100=4(v+v0)/2

Which is 50=v+v0

v=v0 since horizontal acceleration always equals zero

so 2v0 = 50

v0 = 25

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Ahsoka pulls a wounded soldier (mass 64 kg) across the ground with a force 18 Newtons. If they have an acceleration of 1.5 m/s2
Kryger [21]

Answer:

0.124

Explanation:

18-(9.81*64)*k=64*1.5

4 0
3 years ago
Read 2 more answers
What is the internal energy of 3.00 mol of N2 gas at 25 degrees C? To solve this problem, use the equation: U electronic gas = 5
allochka39001 [22]

<u>Given data</u>

Determine Internal energy of gas N₂,  (U) = ?

Temperature (T) = 25° C

                          = 25+273 = 298 K,

Gas constant (R) = 8.31 J/ mol-K ,

Number of moles (n) = 3 moles,

<u>Internal energy of N₂ </u>

Internal energy is a property of thermodynamics, the concept of internal energy can be understand by ideal gas. For example N₂, the observations for oxygen and nitrogen at atmospheric temperatures,  f=5,  (where f is translational  degrees of freedom).

       So per kilogram of gas,

          The internal energy (U) = 5/2 .n.R.T

                                                  = (5/2) × 3 × 8.31 ×298

                                                  = 18572.85 J

<em>The internal energy of the N₂ is 18,572.85 J and it is approximately equal to 18,600 J given in the option B.</em>

 



8 0
3 years ago
Read 2 more answers
Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano
Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

<h2>Part a </h2>

Specific energy is given by

\epsilon=\frac{v^2}{2}-\frac{\mu}{r}

Here

  • ε is the specific energy
  • v is the velocity which is given as 2.23 km/s
  • μ is the gravitational constant whose value is 398600
  • r is the distance between earth and the meteorite which is 402,000 km

                         \epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2

Value of specific energy is also given as

\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km

Orbit formula is given as

r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0

Putting values in this equation and solving for e via the quadratic formula gives

ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

<h2>Part b</h2>

The radius of trajectory at perigee is given as

r_p=a(e-1)\\

Substituting values gives

r_p=133319 (1.086-1)\\r_p=11465.4 km

Now for estimation of altitude z above earth is given as

z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km

So the altitude at closest approach is 5088 km

<h2>Part c</h2>

radius of perigee is also given as

r_p=\frac{h^2}{\mu}\frac{1}{1+e}

Rearranging this equation gives

h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s

Now the velocity at perigee is given as

v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\

So the velocity at perigee is 8.516 km/s

<h2>Part d</h2>

Turn angle is given as

\delta =2 sin^{-1} (\frac{1}{e})

Substituting value in the equation gives

\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08

Aiming radius is given as

\Delta =a \sqrt{e^2-1}

Substituting value in the equation gives

\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km

So the turn angle is 134.08 while the aiming radius is 5641.28 km

3 0
4 years ago
A lamp is 10% efficient.How much electrical energy must be supplied to the lamp each second if it produces 20 J of light energy
k0ka [10]

If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.

20 J/s = 10% of Supply

20 J/s = (0.1) x (Supply)

Divide each side by 0.1:

Supply = (20 J/s) / (0.1)

<em>Supply = 200 J/s  </em>(200 watts)

========================

Here's something to think about:  What could you do to make the lamp more efficient ?  Answer:  Use it for a heater !

If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about.  Suddenly ... bada-boom ... the lamp is 90% efficient !

6 0
4 years ago
A standard light bulb emits light rays in _________ directions.
Ainat [17]
Omni-directional. It means all directional.
4 0
3 years ago
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