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3 years ago

A soccer ball was kicked off the roof of a building going

Physics

1 answer:

WINSTONCH [101]3 years ago

4 0

Answer:

Final vertical velocity = -29m/s

Horizontal distance = 100m

Height = 20.41m

Explanation:

1. The vertical final velocity can be calculated thus: vy = vyo - gt

Where;

vy = vertical velocity (m/s)

vyo = vertical initial velocity (20m/s)

g = acceleration due to gravity (9.8m/s²)

t = time (5s)

Hence, vy = vyo - gt

vy = 20 - (9.8 × 5)

vy = 20 - 49

vy = -29m/s

2. x = V0 x t

Where;

x = horizontal distance (m)

Vo = initial velocity

t = time (s)

x = 20 × 5

x = 100m

3. Maximum height = (voy)²/2g

= 20²/ 2 × 9.8

= 400/19.6

= 20.41m

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Answer:32.24 s

Explanation:

Given

mass of runner (m)=51.8 kg

Constant acceleration(a)= $1.31 m/s^2$

Final velocity (v)=5.47 m/s

Time taken taken to reach 5.47 m/s

v=u+at

$5.47=0+1.31\times t$

$t=\frac{5.47}{1.31}=4.17 s$

Distance traveled during this time is

$s=ut+\frac{1}{2}at^2$

$s=\frac{1}{2}\times 1.31\times 4.17^2=11.42 m$

So remaining distance left to travel with constant velocity=153.57 m

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60,000 meters. no explanation

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(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

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Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of r from a point charge, Q, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$

$PD = 135\times 10^3\text{ V} = 135\text{ kV}$

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be x.

$270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]$

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