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lyudmila [28]
2 years ago
10

In an electricity experiment, a 1.20 g plastic ball is suspended on a 59.0 cm long string and given an electric charge. A charge

d rod brought near the ball exerts a horizontal electrical force F⃗ elec on it, causing the ball to swing out to a 24.0 ∘ angle and remain there.
a. What is the magnitude of Felec ?
b.What is the tension in the string?
Physics
1 answer:
stich3 [128]2 years ago
6 0

Answer:

a.) 5.24 10⁻³ N . b) 0.013 N

Explanation:

a) In absence of other forces, the plastic ball is only subject to the force of gravity (downward) , and to the tension in the string, which are equal each other.

We are told that there exists an horizontal force , of an electric origin, that causes the ball to swing out to a 24º angle (respect the normal) and remain there, so there exists a new equilibrium condition.

In this situation, both the vertical and horizontal components of the external forces acting on the ball (gravity, tension and the electrical force) must be equal to 0.

The only force that has horizontal and vertical components, is the tension in the string.

We can apply Newton's 2nd Law to both directions, as follows:

T cos 24º - mg = 0

-T sin 24º + Fe = 0

where T= Tension in the string.

Fe = Electrical Force

mg = Fg = gravity force

⇒ T = mg/ cos 24º

Replacing in the horizontal forces equation:

-mg/cos 24º . sin 24º = -Fe

∴ Fe = mg. tg 24º = 0.0012 kg. 9.8 m/s². tg 24º = 5.24 10⁻³ N

b) In order to get the value of T, we can simply solve for T the vertical forces component equation , as follows:

T = mg/ cos 24º = 0.0012 kg. 9.8 m/s² / 0.914 = 0.013 N

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Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

v_x = 15 m/s

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

d= v_x t =(15 m/s)(5 s)=75 m

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The weight of a standard object defined as having a mass of exactly 2.9 kg is measured to be 28.449 n. in the same laboratory, a
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2 years ago
A load of 800 N is lifted using a block and tackle having 5 pulleys. If the applied effort is 200 N, calculate
ser-zykov [4K]

Explanation:

Load=800N

Effort=200N

1. Mechanical Advantage = LOAD/EFFORT

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= 4

2 Velocity Ratio = no. Of pulleys =5

3. Efficiency = Mechanical advantage / velocity ratio × 100%

= (4/5)×100%

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