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Rufina [12.5K]
4 years ago
6

A spring-loaded toy gun is used to launch a 12.0-g plastic ball. The spring which has a spring constant of 15.0 N/m is compresse

d 8.00 cm as the ball is pushed into the barrell. When the trigger is pulled the swing is released and shoots the ball back out horizontally. What is the elastic potential energy stored in the spring when it is compressed 8.00 cm?
Physics
1 answer:
ziro4ka [17]4 years ago
4 0

Answer:

the elastic potencial energy stored when the spring is compressed x=8 cm is K(x)= 0.048 J

Explanation:

since the work is related with the force through

W=∫F dx

for a spring of constant k :

F=k*x , where F= compression force, x= compression length

then for a compression from 0 until x

W=∫F dx = ∫ k*x dx =  k ∫x dx =1/2*k*x² - 1/2*k*0² = V(x) - V(0)

since the work depends only on the final value of compression and not on the process 1/2*k*x² represents the elastic potential energy V stored in the spring, then

V(0) = 1/2*k*0² = 0  

V(x) = 1/2*k*x²

when the spring is compressed x= 8 cm = 0.08m , the elastic potencial energy is

V (x) = 1/2*k*x² = 1/2 * 15 N/m* (0.08m)² = 0.048 J

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Answer:

a=12.97\ m/s^2

Explanation:

Given that,

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Speed of the ball, v = 3.36 m/s

We need to find the acceleration of the ball. The acceleration acting on the ball is centripetal acceleration. It is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(3.36)^2}{0.87}\\\\=12.97\ m/s^2

So, the acceleration of the ball is 12.97\ m/s^2.

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3 years ago
A 85 kg lineman tackles a 90 kg receiver. The receiver is running 5.8 m/s, and the lineman is moving 4.1 m/s, at a right angle t
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Answer:

3.59 m/s

Explanation:

We are given that

Mass of lineman,m=85 kg

Mass of receiver,m'=90 kg

Speed of receiver,v'=5.8 m/s

Speed of lineman,v=4.1 m/s

\theta=90^{\circ}

We have to find the their velocity immediately after the tackle.

Initial momentum,P_i=\sqrt{p^2_1+p^2_2}=\sqrt{(85\times 4.1)^2+(90\times 5.8)^2}=627.6 kgm/s

According to law of conservation of momentum

Initial momentum=Final momentum=(m+m')V

627.6=(85+90)V=175V

V=\frac{627.6}{175}=3.59 m/s

3 0
3 years ago
10. Solve the following numerical problems
frosja888 [35]

Answer:

\boxed {\boxed {\sf 120 \ Joules}}

Explanation:

Work is equal to the product of force and distance.

W=F*d

The force is 8 Newtons and the distance is 15 meters.

F= 8 \ N \\d= 15 \ m

Substitute the values into the formula.

W= 8 \ N * 15 \ m

Multiply.

W= 120 \ N*m

  • 1 Newton meter is equal to 1 Joule
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W= 120 \ J

The work done is <u>120 Joules</u>

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3 years ago
A bullet of mass 11 g strikes a ballistic pendulum of mass 1.9 kg. The center of mass of the pendulum rises a vertical distance
Ymorist [56]

Answer:

The bullet's initial speed is 243.21 m/s.

Explanation:

Given that,

Mass of the bullet, m_b=11\ g=0.011\ kg

Mass of the pendulum, m_p=19\ kg

The center of mass of the pendulum rises a vertical distance of 10 cm.

We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,

\dfrac{1}{2}(m_b+m_p)V^2 =(m_b+m_p)gh\\\\V=\sqrt{2gh} \ .................(1)

V is the speed of the bullet and pendulum at the time of collision

Now using conservation of momentum as :

m_bv=(m_p+m_b)V

Put the value of V from equation (1) in above equation as :

v=\dfrac{(m_p+m_b)}{m_b}\sqrt{2gh} \\\\v=\dfrac{(1.9+0.011)}{0.011}\sqrt{2\times 9.8\times 0.1}\\\\v=243.21\ m/s

So, the bullet's initial speed is 243.21 m/s.

7 0
4 years ago
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