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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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Three equal charges of magnitude 'e' are located at the vertices of an equilateral triangle of side 1m. Where should you place a
Naily [24]

Answer:

At centroid

Explanation:

In the given equilateral triangle ABC with side of 1 m. The three equal charges e,e,e are placed at the A,B and C.

And the fourth charge 2e is put at point O which is called centroid.

Now we can calculate the distance AD by applying pythagorean theorem as,

AD^{2}=AB^{2}+BD^{2}

Put the values and get.

AD^{2}=1^{2}+(\frac{1}{2} )^{2}\\AD=\sqrt{\frac{3}{4} } \\AD=\frac{\sqrt{3}}{2}

Now calculate AO as,

AO=\frac{2}{3}\times \frac{\sqrt{3} }{2}\\AO=\frac{1}{\sqrt{3} }

And the sides BO=CO=AO.

Now Force can be calculated as

F_{1}=\frac{2ke^{2} }{\frac{1}{\sqrt{3} } ^{2} }\\F_{1}=6ke^{2}

And similarly,

F_{2}=F_{3}=6ke^{2}

Now we can calculate resultant of F_{2}andF_{3} in upward direction. as,

F_{net}=\sqrt{F_{2}^{2}+F_{3}^{2}+2F_{2}F_{3}cos120  } \\F_{net}=\sqrt{F_{2}^{2}+F_{2}^{2}+2F_{2}F_{2}(-\frac{1}{2})}\\F_{net}=6ke^{2}

Therefore the resultant force on centroid O.

F=F_{1}-F_{net}\\F=6ke^{2}-6ke^{2}\\F=0

Therefore the fourth charge 2e should be placed on centroid so that it experience zero force.

8 0
3 years ago
Match the term to the correct description.
Misha Larkins [42]

1
The flow of electrons from negative to positive.

3
Materials that can easily carry electricity from place to place.

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Materials that do conduct electricity, just not as well as conductors.

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When a dielectric slab is inserted between the plates of one of the two identical capacitors
tatiyna

Answer:(A-P,S;B-P,S;C-Q,S;D-P,S)

Solution

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Explanation:

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Read 2 more answers
At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

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total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

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The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

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