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lidiya [134]
3 years ago
10

If a ball is tossed straight up into the air, at what position is its potential energy the greatest?

Physics
2 answers:
krok68 [10]3 years ago
4 0
A. when it reaches the top of its flight

potential energy increases as the height of the object does, and the highest point of a ball's flight is when it reaches the top. 
abruzzese [7]3 years ago
3 0

Answer:

A.when it reaches the top of its flight

Explanation:

The potential energy of an object is given by:

U=mgh

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth at which the object is located above the ground

Therefore, we see that the higher the object, the larger its potential energy: this means that the ball will have the greatest potential energy at the top of its trajectory.

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What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0
2 years ago
Read 2 more answers
An applied force of 50 N is used to accelerate an object, that weighs 73 N, to the right across a frictional surface. The object
Hunter-Best [27]

Answer:

5.38 m/s^2

Explanation:

NET force causing the object to accelerate  =  50 -10 = 40 N

Mass of the object =  73 N / 9.81 m/s^2 = 7.44 kg

F = ma

40 = 7.44 * a         a = 5.38 m/s^2

6 0
1 year ago
"What is the overall length limitation of a UTP cable run from the telecommunications closet to a networking device in the work
Goryan [66]

Answer:

90-100meters

Explanation:

The overall length limitation of a UTP cable is 90-100meters. Once this limitation is reached, a repeater is employed to transfer data.

4 0
3 years ago
What is the equivalent resistance of a circuit that contains two 50.00
Whitepunk [10]

Answer:

A

Explanation:

Resistors in series add. There is only one path the current can take. That's why Christmas Tree lights sometimes give a lot of trouble. If a bulb burns out, it could be any one of them and time is needed to find the burned out bulb.

That being the case R = R1 + R2

R1 = 50 ohms

R2 = 50 ohms

R = 50 + 50

R = 100 ohms

Answer A

4 0
3 years ago
Read 2 more answers
50 POINTS !!!
shepuryov [24]

Answer:

a) I = 464 kg m²,  b)  K = 631 .6 J, c)  v = 8.25 m / s

Explanation:

a) the moment of inertia of point particles is

          I = ∑ m_i r_i²

in this case

          I = 8 5² + 3 (-2) ² + 7 (-6) ²

          I = 464 kg m²

b) The kinetic energy is

          K = ½ I w²

          K = ½ 464 1.65²

          K = 631 .6 J

c) linear and angular velocity are related

          v = w r

          v = 1.65 5

          v = 8.25 m / s

8 0
2 years ago
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