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grin007 [14]
2 years ago
14

One kind of slingshot consists of a pocket that holds a pebble and is whirled on a circle of radius r. The pebble is released fr

om the circle at the angle θ so that it will hit the target. The angle ϕ in the drawing is 37.1°. The distance to the target from the center of the circle is d. (See the drawing below, which is not to scale.) The circular path is parallel to the ground, and the target lies in the plane of the circle. The distance d is one times the radius r. Ignore the effect of gravity in pulling the stone downward after it is released and find the angle θ.

Physics
1 answer:
Debora [2.8K]2 years ago
5 0

Answer:

θ=142.9°

Explanation:

d=1 *r

angle ϕ= 37.1°

the line connecting pebble and target should be tangent to a circle so

cos(180-ϕ-θ)=\frac{r}{d}=\frac{1}{1}

∴ θ=180-ϕ-cos^{-1} (\frac{1}{1} )

  θ= 180-37.1-0

  θ=142.9°

 

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One of the great dangers to mountain climbers is an avalanche, in which a large mass of snow and ice breaks loose and goes on an
sammy [17]

Answer:

Explanation:

The acceleration of an object down a slope (neglecting friction, µ = 0) is:

a = g × sin θ

Where,

g is the acceleration due to gravity and θ is the angle of the slope.

a = (9.8 × sin (21.5º)

= 3.592 m/s²

Using equations of motion,

S = ut + 1/2at²

Since, u = 0,

S = 1/2at²

347 = 1/2 × (3.592)t²

t² = 193.21

= sqrt(193.21)

= 13.9 s.

5 0
2 years ago
A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A
cluponka [151]

Answer:

D) -Q

Explanation:

The charge inserted will induce -Q charge on the inner surface and + Q on the outer surface of the shell . This charge is called bound charge because it remained attached with opposite charge inserted inside.

5 0
2 years ago
A small crack occurs at the base of a 15.0-m-high dam. The effective area through which water leaves is 2.30 × 10-3 m2. (a) Igno
vova2212 [387]

Answer

given,                                              

height of the dam = 15 m            

effective area of water = 2.3 x 10⁻³ m²

Using energy conservation              

    m g h = \dfrac{1}{2}mv^2

    v= \sqrt{2gh}                  

    v= \sqrt{2\times 9.8 \times 15}

    v= \sqrt{294}              

           v = 17.15 m/s            

 discharge of water

      Q = A V                            

      Q = 2.3 x 10⁻³ x 17.15    

      Q = 0.039 m³/s

3 0
2 years ago
Why is the transverse spatial extent of a photon proportional to its wavelength which is a longitudinal quantity?
otez555 [7]
In quantum mechanics, particularly the wave-particle theory, it states that light behaves like a wave or a particle. For the wave behavior, its movement is measured in wavelengths while the time for each wavelength is the frequency. For the particle behavior, according to Planck, the energy of the photon (light particle) is determined as

E = hc/wavelength, where h is the Planck's constant (<span>6.626 x 10-34 J-s per particle) and c is the speed of light ( 3 x 10^8m/s)

As you can see, the energy of the photon is INVERSELY PROPORTIONAL to the wavelength with the Planck's constant as the constant of proportionality.</span>
7 0
2 years ago
A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen
Svetllana [295]

Answer:

h_p = 30.46\ m

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

v_f=g\cdot t

Where g = 9.8 m/s^2

Similarly, the distance y the object has traveled is calculated as follows:

\displaystyle y=\frac{g\cdot t^2}{2}

If we know the height h from which the object was dropped, we can solve the above equation for t:

\displaystyle t=\sqrt{\frac{2\cdot y}{g}}

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

t_2=2.56 - 2 = 0.56\ sec

Therefore, it has traveled down a distance:

\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m

Thus, the height of the pen is:

h_p = 32 - 1.54\Rightarrow h_p=30.46\ m

8 0
2 years ago
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