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andreyandreev [35.5K]
1 year ago
6

A spring driven dart gun propels a 10g dart. It is cocked by exerting a force of 20N over a distance of 5cm. With what speed wil

l the dart leave the gun, assuming the spring has negligible mass?

Physics
2 answers:
adelina 88 [10]1 year ago
6 0
<span>14 m/s Assuming that all of the energy stored in the spring is transferred to dart, we have 2 equations to take into consideration. 1. How much energy is stored in the spring? 2. How fast will the dart travel with that amount of energy. As for the energy stored, that's a simple matter of multiplication. So: 20 N * 0.05 m = 1 Nm = 1 J For the second part, the energy of a moving object is expressed as KE = 0.5 mv^2 where KE = Kinetic energy m = mass v = velocity Since we now know the energy (in Joules) and mass of the dart, we can substitute the known values and solve for v. So KE = 0.5 mv^2 1 J = 0.5 0.010 kg * v^2 1 kg*m^2/s^2 = 0.005 kg * v^2 200 m^2/s^2 = v^2 14.14213562 m/s = v So the dart will have a velocity of 14 m/s after rounding to 2 significant figures.</span>
kolezko [41]1 year ago
5 0

The dart will leave the gun with speed 10 m/s

\texttt{ }

<h3>Further explanation</h3>

<em>Hooke's Law states that the length of a spring is directly proportional to the force acting on the spring.</em>

\boxed {F = k \times \Delta x}

<em>F = Force ( N )</em>

<em>k = Spring Constant ( N/m )</em>

<em>Δx = Extension ( m )</em>

\texttt{ }

The formula for finding Young's Modulus is as follows:

\boxed {E = \frac{F / A}{\Delta x / x_o}}

<em>E = Young's Modulus ( N/m² )</em>

<em>F = Force ( N )</em>

<em>A = Cross-Sectional Area ( m² )</em>

<em>Δx = Extension ( m )</em>

<em>x = Initial Length ( m )</em>

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

compression of the spring = x = 5 cm = 0.05 m

magnitude of the force = F = 20 N

mass of dart = m = 10 g = 0.01 kg

<u>Asked:</u>

speed of the dart = v = ?

<u>Solution:</u>

<em>We will use </em><em>Conservation of Energy </em><em>formula to solve this problem as follows:</em>

Ep = Ek

\frac{1}{2} F x = \frac{1}{2} m v^2

F x = m v^2

v^2 = F x \div m

v = \sqrt { F x \div m }

v = \sqrt { 20 \times 0.05 \div 0.01 }

v = \sqrt { 100 }

v = 10 \texttt{ m/s}

\texttt{ }

<h3>Learn more</h3>
  • Young's modulus : brainly.com/question/6864866
  • Young's modulus for aluminum : brainly.com/question/7282579
  • Young's modulus of wire : brainly.com/question/9755626

\texttt{ }

<h3>Answer details</h3>

Grade: College

Subject: Physics

Chapter: Elasticity

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Explanation:

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1 year ago
Two identical ladders are 3.0 m long and weigh 600 N each. They are connected by a hinge at the top and are held together by a h
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Answer:

The tension in the rope is 281.60 N.

Explanation:

Given that,

Length = 3.0 m

Weight = 600 N

Distance = 1.0 m

Angle = 60°

Consider half of the ladder,

let tension be T, normal reaction force at ground be F, vertical reaction at top hinge be Y and horizontal reaction force be X.

Y+F=600....(I)

X=T.....(II)

On taking moment about base

X\times l\cos\theta+Y\times l\sin\theta-F\dfrac{l}{2}\sin\theta-T\times d=0

Put the value into the formula

X\times3\cos30+Y\times3\sin30-600\times1.5\sin30-T\times1=0

3\cos30 T-T=600\times1.5\sin30-Y \times3\sin30

1.598T=450-1.5(600-F)....(III)

We need to calculate the force for ladder

2F=600\trimes  2

F=600\ N

We need to calculate the tension in the rope

From equation (3)

1.598T=450-1.5(600-600)

1.598T=450

T=\dfrac{450}{1.598}

T=281.60\ N

Hence, The tension in the rope is 281.60 N.

7 0
2 years ago
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You set a tuning fork into vibration at a frequency of 723 Hz and then drop it off the roof of the Physics building where the ac
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Answer:

Explanation:

Given

Original Frequency f=723\ Hz

apparent Frequency f'=697\ Hz

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

f'=f\cdot \frac{v-v_o}{v+v_s}

where  

f'=apparent frequency  

v=velocity of sound in the given media

v_s=velocity of source

v_0=velocity of observer  

here v_0=0

697=723\cdot (\frac{343-0}{343+v_s})

v_s=(\frac{f}{f'}-1)v

v_s=(\frac{723}{697}-1)\cdot 343

v_s=12.79\approx 12.8\ m/s

i.e.fork acquired a velocity of 12.8 m/s

distance traveled by fork is given by

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v_s^2-0=2\times 9.8\times s

s=\frac{12.8^2}{2\times 9.8}

s=8.35\ m

                                       

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1 year ago
A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is
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Answer:  A) 6.38(10)^{6} m

Explanation:

The equation for the moment of inertia I of a sphere is:

I=\frac{2}{5}mr^{2} (1)

Where:

I=9.74(10)^{37}kg m^{2} is the moment of inertia of the planet (assumed with the shape of a sphere)

m=5.98(10)^{24}kg is the mass of the planet

r is the radius of the planet

Isolating r from (1):

r=\sqrt{\frac{5I}{2m}} (2)

Solving:

r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}} (3)

Finally:

r=6381149.077m \approx 6.38(10)^{6} m

Therefore, the correct option is A.

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The question is incomplete. It comes with a set of answer choices.


These are the answer choices:


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Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.


Both Alex and his friend observe it as 10 m/s.


Both Alex and his friend observe it as less than 10 m/s.



Answer: Both Alex and his friend observe it as 10 m/s.


Justification:


1) The speed is relative to the frame of reference.


2) It is said that the both Alex and his friend are standing still.


3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).


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