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ki77a [65]
2 years ago
9

You are target shooting using a toy gun that fires a small ball at a speed of 15 m/s. When the gun is fired at an angle of 31 de

grees above horizontal, the ball hits the bull's-eye of a target at the same height as the gun. Then the target distance is halved.
At what angle must you aim the gun to hit the bull's-eye in its new position? (Mathematically there are two solutions to this problem; the physically reasonable answer is the smaller of the two.)

\Theta 2=
Physics
1 answer:
Ray Of Light [21]2 years ago
3 0

Answer: 13.1° from the horizontal

Explanation: For projectile

Horizontal distance R of a projectile is:

R= U×Usin2x/g

Where U is the speed of projectile, x is angle of projectile and g= acceleration due to gravity=9.8m/s

R= 15×15sin(2×31)/9.8= 225sin(62)/9.8=20.27m

Therefore if R is halved.

R/2 = 20.27/2=10.14m

Hence the angle for this case is.

Making sin(2x) the subject of formula

Sin2x= Rg/U×U

R is now 10.14 in this case.

Sin2x= 10.14×9.8/15×15=0.4415

2x=arcsin0.4415=26.2

x= 26.2/2

x= 13.1°

Good luck...

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The capacitors in each circuit are fully charged before the switch is closed. Rank, from longest to shortest, the length of time
IgorLugansk [536]

Answer:

C has the greatest capacitance and the lightest load. It will provide current to the load for the longest.

A & E are the same

B is next, and finally

D has the smallest equivalent capacitance and the heaviest load. it will run out of juice the quickest.

Curious that the pictures are out of alphabetic order A B C E D.

Explanation:

6 0
2 years ago
g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the
sdas [7]

Answer:

The vertical distance is  d = \frac{2}{k} *[mg + f]

Explanation:

From the question we are told that

   The mass of the cylinder is  m

    The kinetic frictional force is  f

Generally from the work energy theorem

    E  =  P +  W_f

Here E the the energy of the spring which is increasing and this is mathematically represented as

       E =  \frac{1}{2} * k  *  d^2

Here k is the spring constant

        P is the potential energy of the cylinder which is mathematically represented as

     P  = mgd

And

     W_f  is the workdone by friction which is mathematically represented as

      W_f  =  f *  d

So

    \frac{1}{2} * k  *  d^2 =  mgd +  f *  d

=>    \frac{1}{2} * k  *  d^2 =  d[mg +  f    ]

=>  \frac{1}{2} * k  *  d =  [mg +  f    ]

=> d = \frac{2}{k} *[mg + f]

5 0
2 years ago
At 1 atm pressure, the heat of sublimation of gallium is 277 kj/mol and the heat of vaporization is 271 kj/mol. to the correct n
Andreyy89
Below are the choices that can be found elsewhere:

 a. 268 kJ 
<span>b. 271 kJ </span>
<span>c. 9 kJ </span>
<span>d. 6 kJ
</span>
So the key thing to realize here is what the information given to you actually means. Sublimation is going from a sold to a gas. Vaporization is going from a liquid to a gas. Hence you can create two equations from the information that you have: 

<span>Ga (s) --> Ga (g) delta H = 277 kJ/mol </span>

<span>Ga (l) --> Ga (g) delta H = 271 kJ/mol </span>

<span>From these two equations, you can then infer how to get the melting equation be simply finding the difference between the sublimation (two steps) and vaporization (one step). </span>

<span>Ga (s) --> Ga (l) delta H = 6 kJ/mol </span>

<span>At this point, all you need to do is a bit of stoichiometry. You start with 1.50 mol and multiply by the amount of energy per mole (6 kJ/mol). </span>

<span>*ANSWER* </span>
<span>9 kJ/mol (C)</span>
7 0
2 years ago
An athlete stretches a spring an extra 40.0 cm beyond its initial length. how much energy has he transferred to the spring, if t
marissa [1.9K]
The energy transferred to the spring is given by:
U= \frac{1}{2}kx^2
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k is the spring constant
x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
k=52.9 N/cm = 5290 N/m
x=40.0 cm=0.4 m

so now we can use these data inside the equation ,to find the energy transferred to the spring:
U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J
4 0
2 years ago
1. A 930-kg car traveling 56 km/h comes to a complete stop in 2.0 s. What is the
Juli2301 [7.4K]

The force exerted on the car during this stop is 6975N

<u>Explanation:</u>

Given-

Mass, m = 930kg

Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s

Time, t = 2s

Force, F = ?

F = m X a

F = m X s/t

F = 930 X 15/2

F = 6975N

Therefore, the force exerted on the car during this stop is 6975N

6 0
2 years ago
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