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blagie [28]
4 years ago
10

The barrel of a rifle has a length of 0.855 m. A bullet leaves the muzzle of a rifle with a speed of 553 m/s. What is the accele

ration of the bullet while in the barrel? A bullet in a rifle barrel does not have constant acceleration, but constant acceleration is to be assumed for this problem. Answer in units of m/s 2 .
Physics
1 answer:
Novay_Z [31]4 years ago
5 0

Answer:

Acceleration of the bullet will be 1778835.6 m/sec^2      

Explanation:

We have given length of the barrel refile s= 0.855 m

When the bullet leaves the muzzle its velocity is 553 m/sec

So final velocity v = 553 m/sec

Initial velocity will be 0 that is u = 0 m/sec

According to third equation of motion v^2=u^2+2as

553^2=0^2+2\times a\times 0.855

a=178835.6m/sec^2

You might be interested in
A 97.3 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.67 r
Yuliya22 [10]

Answer:

w = 1.14 rad / s

Explanation:

This is an angular momentum exercise. Let's define a system formed by the three bodies, the platform, the bananas and the monkey, in such a way that the torques during the collision have been internal and the angular momentum is preserved.

Initial instant. The platform alone

        L₀ = I w₀

Final moment. When the bananas are on the shelf

we approximate the bananas as a point load and the distance is indicated

x = 0.45m

        L_f = (m x² + I ) w₁

angular momentum is conserved

         L₀ = L_f

         I w₀ = (m x² + I) w₁

         w₁ = \frac{I}{m x^2 + I}  \ w_o

Let's repeat for the platform with the bananas and the monkey is the one that falls for x₂ = 1.73 m

initial instant. The platform and bananas alone

        L₀ = I₁ w₁

         I₁ = (m x² + I)

           

final instant. After the crash

        L_f = I w

        L_f = (I₁ + M x₂²) w

the moment is preserved

        L₀ = L_f

        (m x² + I) w₁ = ((m x² + I) + M x₂²) w

         (m x² + I) w₁ = (I + m x² + M x₂²) w

we substitute

         w =  \frac{m x^2 +I}{I + m x^2 + M x_2^2} \ \frac{I}{m x^2 + I} \ w_o

         w =  \frac{I}{I + m x^2 + M x_2^2} \ w_o

the moment of inertia of a circular disk is

         I = ½ m_p x₂²

we substitute

         w =  \frac{ \frac{1}{2} m_p x_2^2 }{ \frac{1}{2} m_p x_2^2 + M x_2^2 + m x^2} \ \ w_o

let's calculate

          w =\frac{ \frac{1}{2} \ 97.3 \ 1.73^2 }{ \frac{1}{2} \ 97.3 \ 1.73^2 + 21.9 \ 1.73^2 + 9.67 \ 0.45^2 } \ \ 1.67

          w =  \frac{145.60 }{145.60 \ + 65.54 \ + 1.958} \ \ 1.67

          w = 1.14 rad / s

5 0
3 years ago
Name two energy transformations that occur as Adeline pedals her bicycle up a steep hill and then coasts down the other side.
Juliette [100K]

Potential energy is first transformed into kinetic energy as she pedals, then gravitational as she coasts down the hill.

6 0
4 years ago
Read 2 more answers
A 1000 kg satellite and a 2000 kg satellite follow exactly the same orbit around the earth. What is the ratio F1/F2 of the gravi
Tamiku [17]

Answer:

the <em>ratio F1/F2 = 1/2</em>

the <em>ratio a1/a2 = 1</em>

Explanation:

The force that both satellites experience is:

F1 = G M_e m1 / r²       and

F2 = G M_e m2 / r²

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • r is the orbital radius
  • M_e is the mass of Earth

Therefore,

F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]

F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]

F1/F2 = m1/m2

F1/F2 = 1000/2000

<em>F1/F2 = 1/2</em>

The other force that the two satellites experience is the centripetal force. Therefore,

F1c = m1 v² / r    and

F2c = m2 v² / r

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • v is the orbital velocity
  • r is the orbital velocity

Thus,

a1 = v² / r ⇒ v² = r a1    and

a2 = v² / r ⇒ v² = r a2

Therefore,

F1c = m1 a1 r / r = m1 a1

F2c = m2 a2 r / r = m2 a2

In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,

F1 = F1c

G M_e m1 / r² = m1 a1

a1 = G M_e / r²

also

a2 = G M_e / r²

Thus,

a1/a2 = [G M_e / r²] / [G M_e / r²]

<em>a1/a2 = 1</em>

4 0
3 years ago
A light spring of constant 179 N/m rests vertically on the bottom of a large beaker of water. A 5.32 kg block of wood of density
Digiron [165]

Answer:

Compression of the spring: 0.18 m (downward)

Explanation:

The forces acting on the block of wood are:

- The force of gravity, acting downward, of magnitude mg, where m = 5.32 kg is the mass of the block and g=9.8 m/s^2 is the acceleration due to gravity

- The force exerted by the spring, downward, of magnitude kx, where k=179N/m is the spring constant and x is the elongation of the spring

- The buoyant force, upward, of magnitude \rho V g, where \rho=1000 kg/m^3 is the water density and V the volume of the block

Since the block is in equilibrium, the net force is zero, so we can write

mg+kx-\rho V g=0 (1)

We have to find the volume of the block first. We have:

m = 5.32 kg (mass)

\rho_w = 622 kg/m^3 (wood density)

So, the volume is

V=\frac{m}{\rho_w}=\frac{5.32}{622}=0.0086 m^3

So now we can re-arrange eq.(1) to find the elongation of the spring, x:

x=\frac{-mg+\rho Vg}{k}=\frac{-(5.32)(9.8)+(1000)(0.0086)(9.8)}{179}=0.18 m

So, the spring is compressed by 0.18 m.

7 0
3 years ago
Please explain how to solve 17 and 18
andriy [413]

Answer:

17. h = l − l cos θ

18. 1.40 m

Explanation:

Let's call d the height of the triangle.  We can then say:

h = l − d

Using trig, we can write d in terms of l and θ:

d = l cos θ

h = l − l cos θ

If l = 6 m and l cos θ = 40°:

h = 6 − 6 cos 40

h ≈ 1.40

3 0
3 years ago
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