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alexdok [17]
3 years ago
10

A 0.075-kg arrow is fired horizontally. The bowstring exerts a force on the arrow over a distance of 0.90 m. the bow at 40 m/s.

What average force does the bow apply to arrow?​
Physics
1 answer:
Studentka2010 [4]3 years ago
5 0

Answer:

Use work-energy to solve this

The work done by on the bowstring will be converted into kinetic energy of the arrow

so W = F*d = K = 1/2m*v^2

o v = sqrt(2*F*d/m) = sqrt(2*65N*0.90m/0.075kg) = 39.5 m/s

Explanation:

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Un vagón de carga de 5000-kg se mueve a 2m/s e impacta a un vagón de 10000-kg que se encuentra en reposo. Después de la colisión
Mariulka [41]

Answer:

0.67 m/s

Explanation:

Mass of car 1, m₁ = 5000 kg

Mass of car 2, m₂ = 10,000 kg

Initial speed of car 1, u₁ = 2 m/s

Final speed of car 2, u₂ = 0 (at rest)

We need to find the final velocity of both cars when inelastic collision occurs. The momentum will remain conserved in case of inelastic collision. Using the conservation of momentum. Let V is the final speed.

m_1u_1+m_2u_2=(m_1+m_2)V\\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{5000\times 2}{(5000+10000)}\\\\V=0.67\ m/s

So, after the inelastic collision, they will move with a speed of 0.67 m/s.

6 0
3 years ago
Place several E-Field Sensors at a few points on different equipotential lines, and look at the relationship between the electri
Vlada [557]

Answer:

2.

Explanation:

  • If a charge is moved along a equipotential line, no work is done on the charge.
  • If we remember that the work done by an external force, is just the product of the component of the force parallel to the displacement, if the force produces no work, this means that is perpendicular to the displacement.
  • So, as the electric field is just the force per unit charge, and has the same direction as the force (for a positive charge), it must be perpendicular to any equipotential line.
  • As the electric field (by convention) has the same direction as it would be taken by a positive test charge, and positive charges move from higher voltages to lower ones, the electric field is directed toward lines of lower voltages (like it happens between the plates of a capacitor).
3 0
3 years ago
A diffusion couple composed of two silver– gold alloys is formed; these alloys have compositions of 98 wt% Ag–2 wt% Au and 95 wt
jeyben [28]

Answer:

for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)

It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.

we have D= D0exp( -Qd/RT)

=(8.5×105m2/s)exp(-202,100/8.31×1023)

= 4.03 ×10-15m2/s

4 0
4 years ago
Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1
Artemon [7]

Answer:

\frac{F_1}{F_2} =\frac{1}{2}

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

F_1=2k\times d

case 2:

F_2=2k\times 2d

therefore, \frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}

\frac{F_1}{F_2} =\frac{1}{2}

4 0
4 years ago
While chatting with a friend you place your book bag on a nearby slide in the playground at school. The bag remains stationary.
alukav5142 [94]

Answer:

3. fs < μmg

4. fs = mg sinθ

Explanation:

For any object placed on a slide, there are 3 external forces acting on it:

  • Fg = m*g (always downward)
  • N (normal force, always perpendicular to the surface of the slide. going upward)
  • Fs (Friction Force, always opposite to the movement of the object, parallel to the slide)

As we have only one force with components along the normal and parallel to the slide directions (gravity force), it is advisable to find the components of  this force, along these directions.

If θ is the angle of the slide above the horizontal, we have the following components of Fg:

Fgn = m*g*cosθ

Fgp = m*g*sin θ

We can apply Newton's 2nd Law to these perpendicular directions:

Fp = m*g*sin θ - Fs

Fn = N -m*g*cosθ = 0 (as the object has no movement in the direction perpendicular to the slide) (1)

Looking at the equation for the parallel direction, we have two forces, the component of Fg along the slide (which tries to accelerate the object towards the bottom of the slide), and the friction force.

While the object remains stationary, the equation for Newton's 2nd Law along this direction is as follows:

m*g*sin θ - fs =0 ⇒ fs = m*g*sinθ  (4.)

This force can take any value (depending on the angle θ) to equilibrate the component of Fg along the slide, up to a limit value, which  is given by the following expression:

fsmax = μN (2)

From (1), N= m*g*cos θ

Replacing in (2):

fsmax = μ*m*g*cos θ

While the bag remains at rest, we can say:

fs < μ*m*g*cosθ < μ*m*g (as in the limit cosθ =1)

So, the following is always true:

fs < μmg (3.)

6 0
3 years ago
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