A static charge that is produced by a nearby objet is an?

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s (~ 18 mph). What is the total time it is i

evablogger [386]

Answer:

The total time it is in the air for the ball is 1.6326 s

Given:

Initial velocity = 8 $\frac{m}{s}$

To find:

the total time it is in the air = ?

Formula used:

t = $\frac{v-u}{a}$

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

Solution:

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.

The time taken by the ball to reach the maximum height is given by,

t = $\frac{v-u}{a}$

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

t = $\frac{0-8}{-9.8}$

t = 0.8163 s

Thus, time taken by the ball to reach the ground again = time taken to reach maximum height

So, Total time required for ball to reach ground = 2t = 2 × 0.8163

Total time required for ball to reach ground = 1.6326 s

The total time it is in the air for the ball is 1.6326 s

4 0

3 years ago

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

Dmitrij [34]

Answer:

$\lambda_2 = 573.3 nm$

Explanation:

As we know that the position of maximum intensity on the screen is given as

$y = \frac{N\lambda L}{d}$

here we know that

$\lambda$ = wavelength

L = distance of the screen

d = distance between two slits

now we know that the position of 8th maximum intensity is same as that of 9th maximum on the screen

so we have

$\frac{N_1\lambda_1 L}{d} = \frac{N_2 \lambda_2 L}{d}$

so here we have

$8 (645 nm) = 9 \lambda_2$

$\lambda_2 = 573.3 nm$

4 0

3 years ago

How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00

laiz [17]

Answer:

Work= -7.68×10⁻¹⁴J

Explanation:

Given data

q₁=q₂=1.6×10⁻¹⁹C

r₁=2.00×10⁻¹⁰m

r₂=3.00×10⁻¹⁵m

To find

Work

Solution

The work done on the charge is equal to difference in potential energy

W=ΔU

$Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\ Work=-7.68*10^{-14}J$

4 0

3 years ago

why can scientists ignore the forces of attraction between particles in a gas under ordinary conditions?

Anna35 [415]

<span>The particles in a gas are apart and moving fast, so the forces of attraction are too weak to have a noticeable effect.</span>

5 0

3 years ago

Read 2 more answers

Melting, freezing, and boiling are _____________ changes. Question 3 options: A. compound

vfiekz [6]

Melting freezing and boiling are molecular changes

3 0

3 years ago

Read 2 more answers