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arlik [135]
3 years ago
5

The student soon loses his balance and falls backwards off the board at a velocity of 1.0 m/s. Assuming momentum is conserved in

this process, what is the skateboard's new velocity in meters per second
Physics
1 answer:
Phantasy [73]3 years ago
5 0

Answer:

v2 = 27.3m/s

Explanation:

Assuming forward as positive.

Mass = m1 = 64kg

Let v be the common velocity of the student and the skateboard.

mass of skateboard = m2 = 5.94kg

v = 1.4m/s

Since the skateboard and the student are initially moving together at the same velocity their momentum together is

(m1 + m2)v

Let the final velocity of the student be v1 and the final velocity of the skateboard be v2

v1 = – 1.0m/s (falls backwards that's why the velocity is negative since we are assuming forward as positive)

Then from conservation of momentum, momentum before is equal to momentum after.

(m1 + m2)v = m1v1 + m2v2

m2v2= (m1 + m2)v – m1v1

v2 = ( (m1 + m2)v – m1v1)/m2

v2 = ( (64 + 5.94)×1.4 – 64×(-1.0))/5.94

v2 = ( (64 + 5.94)×1.4 + 64×1.0)/5.94

v2 = 27.3m/s

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Cart A, with a mass of 0.20 kg, travels on a horizontal air trackat 3.0m/s and hits cart B, which has a mass of 0.40 kg and is i
inysia [295]

Answer:

\large \boxed{\text{C. 2.3 m/s}}

Explanation:

Data:

m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\

Calculation:

This is a perfectly inelastic collision.  The two carts stick together after the collision and move with a common final velocity.

The conservation of momentum equation is

\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}}  + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}

3 0
3 years ago
A 15g bullet is fired horizontally into a 3kg block of wood suspended by a long cord. Assume that the bullet remains in the bloc
Varvara68 [4.7K]

Answer:

261.3 m/s

Explanation:

Mass of bullet=m=15 g=\frac{15}{1000}=0.015 kg

1 kg=1000g

Mass of block=M=3 kg

d=0.086 m

Total mass =M+m=3+0.015=3.015 kg

K.E at the time strike=Gravitational potential energy at the end of swing

\frac{1}{2}(m+M)^2V^2=(m+M)gh

Using g=9.8m/s^2

Substitute the values

\frac{1}{2}(3.015)V^2=3.015\times 9.8\times 0.086

V^2=\frac{2\times 3.015\times 9.8\times 0.086}{3.015}

V=\sqrt{2\times 3.015\times 9.8\times 0.086}}

V=1.3m/s

Velocity after collision=V=1.3 m/s

Velocity of block=v'=0

Using conservation law of momentum

mv+Mv'=(m+M)V

Using the formula

0.015v+3(0)=3.015(1.3)

0.015v=3.015(1.3)

v^2=\frac{3.015(1.3)}{0.015}=261.3

v=261.3 m/s

5 0
3 years ago
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b
Lesechka [4]

Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

        C = Q / V

        Q = C V

         

can also be calculated using geometry consideration

        C = e or A / d

         

we reduce to the SI system

       A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

       d = 1.53 cm = 1.53 10⁻² m

we substitute

         Q = eo A / d V

         Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

         Q = 3.9757 10⁻¹⁰ C

         

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

           V = Vo / k

            V = 275/80

            V = 3.4375 V

c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

             U = 5.47 10⁻⁸ J

let's reduce to nJ

              109 nJ = 1 J

               U = 54.7 nJ

d) energy after submerging

             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

             U = U₀ / k

             U = 54.7 / 80 nJ

              U = 0.68375 nJ

the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

           

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Answer:

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