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Cloud [144]
2 years ago
10

8. A 3.0 x 10' g nerf projectile is fired from a 1.5 kg nerf gun. If the velocity of the projectile is

Physics
1 answer:
tatuchka [14]2 years ago
4 0

Answer:

v_g=0.2\ m.s^{-1}

Explanation:

Given:

  • mass of nerf projectile, m_p=30\ g=0.03\ kg
  • velocity of nerf projectile, v_p=10\ m.s^{-1}
  • mass of the gun, m_g=1.5\ kg

Now in this case the collision can be assumed to be perfectly elastic.

<u>For an elastic collision:</u>

m_p.v_p=m_g.v_g

where:

v_g= recoil velocity of the gun

0.03\times 10=1.5\times v_g

v_g=0.2\ m.s^{-1}

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A 65-kg bungee jumper, who is attached to one end of an 85-m long bungee cord that has its other end tied to a bridge, jumps off
irinina [24]

Answer:

The impulse delivered to the bungee jumper is 1.32 kN.s

Explanation:

The situation can be shown graphically as shown in the figure.

Impulse delivered to the bungee jumper = Area under the curve.

The curve represents a triangle and the area of traiangle = (1/2)base×height

The base of the triangle from the graph = 1.2 seconds.

The height of the triangle from the graph = 2.2 kN

Thus,

<u>Impulse = (1/2)×(1.2 seconds)×(2.2 kN) = 1.32 kN.s</u>

3 0
3 years ago
You are given two vectors vector A = 4.9 at 31o vector B = 6 at 156o Angles are measured counterclockwise from the x-axis. What
Ket [755]

Answer:

   C_{y} = 4.96  and     θ' = 104,5º

Explanation:

To add several vectors we can decompose each one of them, perform the sum on each axis, to find the components of the resultant and then find the module and direction.

Let's start by decomposing the two vectors.

Vector A

             sin θ = A_{y} / A

             cos θ = Aₓ / A

             A_{y} = A sin  θ

             Ax = A cos θ

             A_{y} = 4.9 sin 31 = 2.52

             Ax = 4.9 cos 31 = 4.20

Vector B

           B_{y} = B sin θ

           Bx = B cos θ

           B_{y} = 6 sin 156 = 2.44

           Bx = 6 cos 156 = -5.48

The components of the resulting vector are

X axis

         Cx = Ax + B x

         Cx = 4.20 -5.48

         Cx = -1.28

Axis y

         C_{y} = Ay + By

         C_{y} = 2.52 + 2.44  

         C_{y} = 4.96

Let's use the Pythagorean theorem to find modulo

         C = √ (Cₙ²x2 + Cy2)

         C = Ra (1.28 2 + 4.96 2)

         C = 5.12

We use trigonemetry to find the angle

         tan θ = C_{y} / Cₓ

          θ’ = tan⁻¹ (4.96 / (1.28))

           θ’ = 75.5

como el valor de Cy es positivo y Cx es negativo el angulo este en el segundo cuadrante, por lo cual el angulo medido respecto de eje x positivo es

       θ’ = 180 – tes

        θ‘= 180 – 75,5

        θ' = 104,5º

7 0
3 years ago
A body accelerate uniformly from rest at 2m/s square.Calculate its velocity after traveling 9m
Thepotemich [5.8K]
The correct answer for this question is 6m/s. I hope this helps.
5 0
2 years ago
If a 15 kg mass accelerates at a rate of 4 m/s2, what net force acts on it?
jarptica [38.1K]
<h2>Answer: D 60N</h2>

<h3>Explanation:</h3>

Mass(M)=15 kg

Acceleration(A)=4 m/s2

Force=?

Now,

Force(F)=M×A

F=15×4

F=60N Ans

5 0
2 years ago
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the
algol13

Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa

<u>Explanation:</u>

Given -

Stress Direction, A = [1 0 0 ]

Slip plane = [ 1 1 1]

Normal to slip plane, B = [ 1 1 1 ]

Critical stress, Sc = 2.92 MPa

Let the direction of slip on = [ 1 1 0 ]

Let Ф be the angle between A and B

cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3

cos Ф = 1/√3

σ = Sc / cosФ cosλ

For slip along [ 1 1 0 ]

cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1

cos λ = 1/√2

Therefore,

σ = 2.92 / 1/√3 1/√2

σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa

Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa

 

4 0
3 years ago
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