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FrozenT [24]
3 years ago
11

The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with

mass 0.0300 kg is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 cm long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal.Calculate the speed with which the ball leaves the barrel if you can ignore friction.
Physics
1 answer:
Molodets [167]3 years ago
5 0

Answer:

v = 6.93 m/s

Explanation:

given,

spring constant = k = 400 N/m

compression of spring = 6 cm = 0.06 m

mass of ball = 0.03 Kg

length of barrel = 6 cm

using energy conservation

KE of ball = PE of spring

\dfrac{1}{2}mv^2 =\dfrac{1}{2}kx^2

mv^2 = kx^2

v = \sqrt{\dfrac{kx^2}{m}}

v = \sqrt{\dfrac{400\times 0.06^2}{0.03}}

v = \sqrt{48}

v = 6.93 m/s

hence, the speed at which ball leaves the barrel will be equal to v = 6.93 m/s

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malfutka [58]

Answer:

a

   \theta  =  23.32^o

b

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c

s =  0.948 \  m

Explanation:

From the question we are told that

The mass of the bag is m_b  =  25.0 \  kg

The normal force experienced is F_n  =  225 \ N

The maximum acceleration of the bag is a =  2.40 \  m/s^2

Generally this normal force experience by the bag is mathematically represented as

F_n  =  mg cos \theta

=> 225  =  (25 * 9.8) cos \theta

=> 0.9183  =   cos \theta

=> \theta  = cos^{-1}[0.9183]

=> \theta  =  23.32^o

Generally for the bag not to slip , it means that the frictional force is equal to the sliding force

F_f =  F_s

Hence F_f is mathematically represented as

F_f   =  \mu_s  *  F_n

While F_s is mathematically represented as

F_s   =  m * a

So

\mu_s  *  F_n = m * a

=> \mu_s  *  225 = 25 * 2.40

=> \mu_s =  0.27

Generally from the workdone equation we have that

KE_f - KE_i =  W_f

Here W_f is the work done by friction which is mathematically represented as

W_f  =  m * g * \mu_k * s

Here s is the distance covered by the bag

KE_f is zero given that velocity at rest is zero

and

KE_i = \frac{1}{2}  *  m* v_i^2

so

   \frac{1}{2}  *  m* v_i^2 = m * g * \mu_k * s

=>  \frac{1}{2}  *  v_i^2 =   g * \mu_k * s

substituting  2.55 m/s for v_i and 0.350 for  \mu_k  we have that

     \frac{1}{2}  *  2.55^2 =   9.8 * 0.350 * s

=> s =  0.948 \  m

4 0
3 years ago
Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant
levacccp [35]

This question is incomplete, the complete question is;

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Answer:

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Explanation:

Given the data in the question and image below and as illustrated in the second image;

distance S = 40 m

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V_A = 72 km/hr

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now, angular velocity of Bxy will be;

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ω_B = ( 54 × 1000/3600) / 100

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Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

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mylen [45]
The first three choices: a, b and c can be considered reconstruction except the last one which is letter d. I'm not really certain what reconstruction is, but judging from the patterns of the first three choices, reconstruction could mean that an inference is made after a part of an event has proved that event to be true. 
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Gemiola [76]

Answer:

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WORK = POWER * TIME

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