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serious [3.7K]
3 years ago
9

Two golf balls are hit from the same point on a flat field. Both are hit at an angle of 55 degree above the horizontal. Ball 2 h

as twice the initial speed of ball 1. If ball 1 lands a distance d_1 from the initial point, at what distance d_2 does ball 2 land from the initial point? (Neglect any effects due to air resistance.)
d_2 = d_1
d_2 = 4d_1
d_2 = 8d_1
d_2 = 0.5d_1
d_2 = 2d_1
Physics
1 answer:
juin [17]3 years ago
4 0

Answer:

d_2 = 4d_1

Explanation:

The range or horizontal distance covered by a projectile projected with a velocity U at an angel of θ to the horizontal is given by

R = U²sin2θ/g

Let the range or horizontal distance of ball 1 with initial velocity U projected at an angle θ = 55° be

d_1 = U²sin2θ/g

Let the range or horizontal distance of ball 2 with initial velocity V = 2U projected at an angle θ = 55° be

d_2 = V²sin2θ/g

= (2U)²sin2θ/g

= 4U²sin2θ/g

= 4d_1   (since d_1 = U²sin2θ/g)

So, the ball 2 lands a distance d_2 = 4d_1 from the initial point.

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Which would take a longer time to warm up, a piece of brass or a piece of marble, given that both
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A missile is fired from a jet flying horizontally at Mach 1 (1100 ft/s). The missile has a horizontal acceleration of 1000 ft/s
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Read 2 more answers
Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is
marshall27 [118]

Answer:

x=0.0049\ m= 4.9\ mm

d=0.01153\ m=11.53\ mm

Explanation:

Given:

  • mass of the object, m=0.75\ kg
  • elastic constant of the connected spring, k=150\ N.m^{-1}
  • coefficient of static friction between the object and the surface, \mu_s=0.1

(a)

Let x be the maximum distance of stretch without moving the mass.

<em>The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.</em>

f=k.x

\mu_s.N=k.x

where:

N = m.g = the normal reaction force acting on the body under steady state.

0.1\times (9.8\times 0.75)=150\times x

x=0.0049\ m= 4.9\ mm

(b)

Now, according to the question:

  • Amplitude of oscillation, A= 0.0098\ m
  • coefficient of kinetic friction between the object and the surface, \mu_k=0.085

Let d be the total distance the object travels before stopping.

<em>Now, the energy stored in the spring due to vibration of amplitude:</em>

U=\frac{1}{2} k.A^2

<u><em>This energy will be equal to the work done by the kinetic friction to stop it.</em></u>

U=F_k.d

\frac{1}{2} k.A^2=\mu_k.N.d

0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d

d=0.01153\ m=11.53\ mm

<em>is the total distance does it travel before stopping.</em>

7 0
3 years ago
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