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igor_vitrenko [27]
3 years ago
11

The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively. If there is no

air resistance, how long does it take the football to reach the top of its trajectory?
Physics
1 answer:
Dmitrij [34]3 years ago
7 0

Answer:

t= 2 sec

Explanation:

Given that

Ui= 16 m/s

Vi= 20 m/s

As we know that at the maximum height ,the y component of velocity will become zero.

We know that horizontal component of velocity will remain constant trough out the motion.

We also know that acceleration due to gravity is in downward direction.

Vf= Vi- g t

Now by putting the values

0 = 20 - 10 x t                ( take g= 10 m/s²)

20 = 10 t

t= 2 sec

Therefore the answer is 2 sec.

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Suppose a uniform electric field 4 N/C is in the positive x-direction .
LuckyWell [14K]

The magnitude of the electric field at x =4m on the x axis at this time 1 N/C.

<h3>Electric field at position 4 m</h3>

Electric field at a given distance is calculated as follows;

E = kq/r²

E₂ = (9 x 10⁹ x q)/(2²)

E₂ = 2.25 x 10⁹q

E₂ + E₀ = 0

2.25 x 10⁹q + 4 = 0

2.25 x 10⁹q = - 4

q = -4 / (2.25 x 10⁹)

q = -1.78 x 10⁻⁹

E₄ = (9 x 10⁹ x (-1.78 x 10⁻⁹) ) / (4²)

E₄ = - 1 N/C

|E₄| = 1 N/C

Thus, the magnitude of the electric field at x =4m on the x axis at this time 1 N/C.

The complete question is below:

Suppose a uniform electric field of 4N/C is in the positive x direction. When a charge is placed and at a fixed to the origin, the resulting electric field on the x axis at x =2m becomes zero. What is the magnitude of the electric field at x =4m on the x axis at this time?

Learn more about electric field here: brainly.com/question/14372859

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How has the jester archetype evolved over thr years?
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Find the minimum thickness (in nm) of a soap bubble that appears green when illuminated by white light perpendicular to its surf
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Explanation:

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A 1.94-m-diameter lead sphere has a mass of 5681 kg. A dust particle rests on the surface. What is the ratio of the gravitationa
scoundrel [369]

To develop this problem we will apply Newton's laws regarding gravitational forces, both in space and on earth. From finding this relationship, leaving the variable of the dust mass open, we will find the relationship of the forces between the two surfaces. Our values are,

\text{Diameter of the lead sphere} =  D=1.940m

\text{Mass of the lead sphere} =  m_1 = 5681kg

\text{Mass of the dust particle} = m_2

Distance between the center of lead sphere to dust particle

r = \frac{D}{2}

r = \frac{1.940m}{2}

r = 0.97m

Gravitational force of the sphere on the dust particle:

F = \frac{Gm_1m_2}{r^2}

F = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5681kg)(m_2)}{(0.97m)^2}

F = (4.027*10^{-7} N/kg)m_2

Weight of the dust particle

W = m_2 g

W = m_2 (9.8m/s^2)

Ratio of F and W:

\frac{F}{W} = \frac{(4.07*10^{-7}N/kg) m_2)}{m_2(9.8m/s^2)}

\frac{F}{W} = 4.153*10^{-8}

Therefore the ratio is 4.153*10^{-8}

3 0
3 years ago
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