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andrey2020 [161]
4 years ago
6

A ball is thrown vertically upwards. It returns 6s later. Calculate : (1) the greatest height reached by the ball, and (2) the i

nitial velocity of the ball. (Take g=9.8m/s2) ​
Physics
1 answer:
dsp734 years ago
6 0

Answer:

greatest displacement = 44.1m

initial velocity= 29.4m/s

Explanation:

Greatest displacement

s=1/2at^2

= (9.8/2 ×9)m

= 44.1m

initial velocity

s=ut-1/2at^2

44.1= 3u -(1/2×9.8×9)

44.1=3u-44.1

3u=88.2

u=29.4m/s

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A guy wire 1034 feet long is attached to the top of a tower. When pulled taut, it touches level ground 699 feet from the base of
kolezko [41]

Answer:

80.386 degrees

Explanation:

We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse

The adjacent side = 699ft

The hypotenuse = 1034ft

using cos∅ = Adjacent/hypotenuse

where ∅ is the unknown angle

cos ∅ = 699/1034 = 0.167

∅ = arccos 0.167 = 80.368°

As easy as one can imagine

8 0
3 years ago
In which phase of matter are the atoms closely packed but still able to slide past each other?
borishaifa [10]
The phase of matter is the Solid Phase.
3 0
4 years ago
To warm 2.0 l of tea (d = 1.01 g/ml; sp. heat = a cook places a 500 g block of stone at a temperature of 200f into the teapot. a
IRISSAK [1]
Volume of tea V = 2.0L = 2000 mL density d = 1.01 g/ mL mass of tea m = V * d = 2000mL * 1.01g/mL = 2020 gWhen we assume that the tea was initially at 72, the final temperature of the tea in F is 91. 
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4 0
4 years ago
The force of gravity is less between two objects that
Aloiza [94]

The force of gravity is less between to objects when the product
of their masses is smaller, or when they are farther apart.


6 0
3 years ago
A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.
tatiyna

Answer:

a)   x = ⅔ d , b) the charge must be negative, c) Q

Explanation:

a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing

         ∑ F = 0

        -F₁₂ + F₂₃ = 0

         F₁₂ = F₂₃

         

let's replace the values

        k Q Q / r₁₂² = k Q 4Q / r₂₃²

            Q² / r₁₂² = 4 Q² / r₂₃²

suppose charge 3 is placed at point x

        r₁₂ = x

        r₂₃ = d-x

             

we substitute

             1 / x² = 4 / (d-x) 2

             1 / x = 2 / (d-x)

             x = 2 (x-d)

             x = 2x -2d

            3x = 2d

              x = ⅔ d

b) The sign of the charge must be negative, to have an attractive charge on the two initial charges

c)  Q

5 0
4 years ago
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