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Cloud [144]
3 years ago
6

A rifle is aimed horizontally at a target 30m away. The bullet hits the target 2.9 cm below the aiming point. What is the bullet

's speed ( in the unit of m/s) as it emerges from the rifle.
Physics
2 answers:
Luden [163]3 years ago
6 0

Answer:

Speed of bullet = 389.61 m/s.

Explanation:

Considering the vertical motion of bullet

Initial vertical speed = 0 m/s

Vertical displacement = 2.9 cm = 0.029 m

Vertical acceleration = 9.81 m/s²

Substituting in s = ut + 0.5at²

    0.029 = 0 x t + 0.5 x 9.81 x t²

    t = 0.077 s

So ball hits the target after 0.077 s.

Now considering the vertical motion of bullet

Time = 0.077 s

Horizontal displacement = 30 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    30 = u  x 0.077 + 0.5 x 0 x 0.077²

     u = 389.61 m/s

Speed of bullet = 389.61 m/s.

Rina8888 [55]3 years ago
5 0

Answer:

390 m/s

Explanation:

Let the horizontal speed be u.

Horizontal distance , x = 30 m

Vertical distance, y = 2.9 cm

Let time taken be t.

Use second equation of motion in vertical direction

H = uy × t + 1/2 gt^2

0.029 = 0.5 × 9.8 × t^2

t = 0.077 s

Horizontal distance = horizontal velocity × time

30 = u × 0.077

u = 390 m/s

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Is the visible light color spectrum a small segment of the electromagnetic spectrum?
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Answer:

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Explanation:

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3 years ago
You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov
Marina CMI [18]

Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is  </em>

<em>    11.62 Hz</em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

f_{1} = f[\frac{v_{s} }{v_{s} -v} ] ..............................1

where f_{1} is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = 36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}

v = 16.27 m/s

Substituting into equation 1 we have

f_{1} =  231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})

f_{1}  = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]

where f_{2} is the frequency of the reflected sound;

f_{1}  is the frequency which the wave strikes the hill = 242.61 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]

f_{2}  = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound  and the reflected sound. This can be obtained as follows;

Δf = f_{2} -  f_{1}  

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf  = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0
3 years ago
Two blocks are connected by a very light string passing over a massless and frictionless pulley. Traveling at constant speed, th
Semenov [28]

Answer:

<h2>i. 9J and </h2><h2>ii.  -9J</h2>

Explanation:

Please see attached the diagram and also the FBD for your reference.

Step one:

given data

Weight w1=20N

Distance s1=75cm= 0.75m

Weight w2=12N

Distance s2=75cm =0.75m

Clearly, we are asked to find the work done by gravity and the tension in the string.

we know that work is defined as force times the distance traveled

W=F*D

and to know the force we need the acceleration, seeing that the system is stationary the acceleration due to gravity 9.81m/s^2 is not needed here

i . Work done by gravity = w2*s2=12*0.75=9J

ii. The tension is equal but opposite = -9J

4 0
3 years ago
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