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finlep [7]
3 years ago
14

Two golf balls are hit from the same point on a flat field. Both are hit at an angle of 20° above the horizontal. Ball 2 has twi

ce the initial speed of ball 1. If ball 1 lands a distance d1 from the initial point, at what distance d2 does ball 2 land from the initial point? (Neglect any effects due to air resistance.)
A) dd_{2} = 2d_{_{1}}

B) d_{2} = 0.5d_{_{1}}

C)d_{2} = 8d_{_{1}}

D) d_{2} = d_{_{1}}

E) d_{2} = 4d_{_{1}}
Physics
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer:

E) d_{2} = 4d_{_{1}}

Explanation:

We know that the range of a projectile is given by:

R=\frac{u^2.sin\ \2\theta}{g}

where:

u = initial velocity of the projectile

\theta = angle of projection of projectile

<u>Now, range for ball 1 :</u>

d_1=\frac{u^2.sin\ \2\theta}{g}

<u>Now, range for ball 2 with double velocity and all the other parameters being the same:</u>

d_2=\frac{(2u)^2.sin\ \2\theta}{g}

d_2=4\times \frac{u^2.sin\ \2\theta}{g}

d_2=4\times d_1

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jasenka [17]

Weight of the carriage =(m+M)g =142.1\ N

Normal force =Fsin(\theta) + W = 197.1\ N

Frictional force =\mu N=27.59\ N

Acceleration =4.66\ m\ s^{-2}

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

  • So Weight(W) =(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with F_x, force of 110\ N acting vertically downward.Both are downward and Normal is upward so Normal force =Summation\ of\ both\ forces

  • Normal force (N) = Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N
  • Frictional force (f) =\mu N=0.14\times 197.1 =27.59\ N

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that F_y=Horizontal and F_x=Vertical component of forces.

So Fnet = Fy(Horizontal) - f(friction) = m\times a

  • Acceleration (a) =\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }

So we have the weight of the carriage, normal force,frictional force and acceleration.

3 0
3 years ago
Depth of a pond seems shallower than real depth,why?​
IgorLugansk [536]

Answer:

The refraction of light at the surface of water makes ponds and swimming pools appear shallower than they really are. A 1m deep pond would only appear to be 0.75 m deep when viewed from directly above. When light emerges from glass or water into air it speeds up again.

Explanation:

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olasank [31]

It is actually caused by the environment, so its false. :)

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10m= (5.0) + (.5)(9.8)(5.0)
Norma-Jean [14]
M=2.45 because you multiply out the equation on the right and divide by 10
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27. The traffic officer issued violation tickets to traffic
Evgen [1.6K]

Answer:

27: 85

28:75%

Explanation:

27:68=80

?=100 hence (68×100)÷80

=85

28:<em>1</em><em>8</em><em>/</em><em>2</em><em>4</em><em>×</em><em> </em><em>1</em><em>0</em><em>0</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>7</em><em>5</em><em>%</em>

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