answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
andreyandreev [35.5K]
2 years 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]2 years 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]2 years 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

You might be interested in
In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a
fgiga [73]

(a) The diffraction decreases

The formula for the diffraction pattern from a single slit is given by:

sin \theta = \frac{n \lambda}{a}

where

\theta is the angle corresponding to nth-minimum in the diffraction pattern, measured from the centre of the pattern

n is the order of the minimum

\lambda is the wavelength

a is the width of the opening

As we see from the formula, the longer the wavelength, the larger the diffraction pattern (because \theta increases). In this problem, since the wavelength of the signal has been decreased from 54 cm to 13 mm, the diffraction of the signal has decreased.

(b) 10.8^{\circ}

The angular spread of the central diffraction maximum is equal to twice the distance between the centre of the pattern and the first minimum, with n=1. Therefore:

sin \theta = \frac{(1) \lambda}{a}

in this case we have

\lambda=54 cm = 0.54 m is the wavelength

a=5.7 m is the width of the opening

Solving the equation, we find

\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.54 m}{5.7 m})=5.4^{\circ}

So the angular spread of the central diffraction maximum is twice this angle:

\theta = 2 \cdot 5.4^{\circ}=10.8^{\circ}

(c) 0.26^{\circ}

Here we can apply the same formula used before, but this time the wavelength of the signal is

\lambda=13 mm=0.013 m

so the angle corresponding to the first minimum is

\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.013 m}{5.7 m})=0.13^{\circ}

So the angular spread of the central diffraction maximum is twice this angle:

\theta = 2 \cdot 0.13^{\circ}=0.26^{\circ}

5 0
2 years ago
A 65 kg students is walking on a slackline, a length of webbing stretched between two trees. the line stretches and sags so that
polet [3.4K]

Answer : Tension in the line = 936.7 N

Explanation :

It is given that,

Mass of student, m = 65 kg

The angle between slackline and horizontal, \theta=20^0

The two forces that acts are :

(i) Tension

(ii) Weight

So, from the figure it is clear that :

mg=2T\ sin\ \theta

T=\dfrac{mg}{2\ sin\theta}

T=\dfrac{65\ kg\times 9.8\ m/s^2}{2(sin\ 20)}

T=936.7\ N

Hence, this is the required solution.

5 0
1 year ago
Read 2 more answers
A ray of light passes from air into carbon disulfide (n = 1.63) at an angle of 28.0 degrees to the normal. what is the refracted
Snezhnost [94]
We can solve the problem by using Snell's law, which states 
n_i \sin \theta_i = n_r \sin \theta_r
where
n_i is the refractive index of the first medium
\theta_i is the angle of incidence
n_r is the refractive index of the second medium
\theta_r is the angle of refraction

In our problem, n_i=1.00 (refractive index of air), \theta_i = 28.0^{\circ} and n_r=1.63 (refractive index of carbon disulfide), therefore we can re-arrange the previous equation to calculate the angle of refraction:
\sin \theta_r =  \frac{n_i}{n_r}  \sin \theta_r =  \frac{1.00}{1.63}  \sin 28.0^{\circ} = 0.288
From which we find
\theta_r = \arcsin (0.288)=16.7^{\circ}
6 0
2 years ago
An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in
vesna_86 [32]

Answer:

height =20m

Explanation:

gpe=mgh

800=4×10×x

40x=800

x=20

3 0
2 years ago
As a youngster, you drive a nail in the trunk of a young tree that is 3 meters tall. The nail is about 1.5 meters from the groun
Lera25 [3.4K]

Answer:

15m

Explanation:

Hello! first to solve this problem we must find that so much that the tree grew in the 15 years this is achieved by dividing the height of the tree before and after

\frac{30m}{3m} =10 times

the tree grew 10 times its initial length in 15 years, so to find how tall the nail is, we multiply this factor by 1.5m

X=(1.5m)(10)=15m

the nail is 15 meters above ground level

8 0
2 years ago
Other questions:
  • Choose which statements correctly identify the relationship of mass volume and density by clicking the sentence
    8·1 answer
  • A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer
    7·1 answer
  • A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington
    5·1 answer
  • A 1500 W radiant heater is constructed to operate at 115 V. (a) What will be the current in the heater? (b) What is the resistan
    12·1 answer
  • A floating ice block is pushed through a displacement d = (14 m) i hat - (11 m) j along a straight embankment by rushing water,
    15·1 answer
  • An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular accele
    10·1 answer
  • A vehicle has an initial velocity of v0 when a tree falls on the roadway a distance xf in front of the vehicle. The driver has a
    9·1 answer
  • dopasuj wartości pracy z ramki do przedstawionych sytuacji ,a nastepnie wyraż te prace w dżulach uwaga jedna wartośc pracy nie b
    15·1 answer
  • To open a door, you apply a force of 10 N on the door knob, directed normal to the plane of the door. The door knob is 0.9 meter
    10·1 answer
  • I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!