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2 years ago

A 400 g ball swings in a vertical cirde at the end of

Physics

1 answer:

velikii [3]2 years ago

8 0

Answer:

15.10m/s

Explanation:

The mass of the ball(m)=400g = 0.4kg

The radius of the string is(r)=15m

The tension in the string is(T)=10N

The acceleration due to gravity = $9.8m/s^{2}$

The tension in the string when the body is at the bottom is given by

$T=\frac{mv^{2} }{r}+mg$

To find the speed of the ball, we make v the subject of the formula

Therefore, $v=\sqrt\frac{r(T-mg}{m}$

$v= \sqrt\frac{15(10-0.4*9.8)}{0.4}$

$v=\sqrt\frac{91.2}{0.4} \\$

$v=\sqrt228 = 15.10 m/s$

The speed of the ball = 15.10m/s

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Dr. John Paul Stapp was U.S. Air Force officer whostudied the effects of extreme deceleration on thehuman body. On December 10,

Ira Lisetskai [31]

Answer:

(a) a = 56.4 m/s², his acceleration a, in multiples of gravity g, is 5.76 g

(b) a = -201.43 m/s², his deceleration -a, in multiples of gravity g, is -20.56 g

Explanation:

(a)

When moving upwards, the initial velocity, u = 0 (he accelerated from rest)

When moving upwards, the final velocity, v = 282 m/s

time of motion during this acceleration, t = 5 s

His acceleration is calculated as;

v = u + at

282 = 0 + 5a

a = 282 / 5

a = 56.4 m/s²

Ratio of his acceleration, a to gravity, g = a/g = 56.4 / 9.8 = 5.76

a = 5.76 g

(b)

When moving downwards, the initial velocity, u = 282 m/s

When moving downwards, the final velocity, v = 0 (he was brought to rest)

time of motion during this deceleration, t = 1.4 s

His deceleration is calculated as;

v = u + at

0 = 282 + 1.4a

1.4a = -282

a = -282 / 1.4

a = -201.43 m/s²

Ratio of his deceleration, -a to gravity, g = -a/g = 201.43 / 9.8 = 20.56

a = -20.56 g

3 0

3 years ago

In a classroom demonstration, students are using a Slinky to observe and learn about wave properties. If the Slinky has a period

Andrej [43]

Answer:

Frequency = 3.0 Hertz

Explanation:

Given the following data;

Period = 0.333 seconds

To find the frequency;

Mathematically, frequency of a wave is given by the formula;

Frequency = 1/period

Substituting into the formula, we have;

Frequency = 1/0.333

Frequency = 3.0 Hertz

3 0

3 years ago

The metal wire in an incandescent lightbulb glows when the lights is switch on and stops glowing when switched off. This simple

lutik1710 [3]

Answer:

When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat.

Explanation:

4 0

2 years ago

Read 2 more answers

Find the ratio of the diameter of iron to copper wire, if they have the same resistance per unit length (as they might in househ

Natasha_Volkova [10]

Answer:

The ratio of the diameter of iron to Cu is;

$\frac{d{Fe} }{ d{Cu} } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}$

Explanation:

R=(ρL)/A

R is resistance,
L is length,
A is area,
ρ is resistivity
d is diameter

from the question the two materials have the same resistance per unit length.

$\frac{R}{L}= \frac{p}{A}$

$\frac{R}{L}$ for iron = $\frac{R}{L}$ for copper

This means we can equate ρ/A for both materials.

$\frac{p_{Fe} }{A_{Fe} } =\frac{p_{Cu} }{A_{Cu} }$

re-arranging the equation we have,

$\frac{A_{Fe}}{A_{Cu} } =\frac{p_{Fe} }{ p_{Cu} }$

$A=\pi \frac{d^{2} }{4}$

$\frac{A_{Fe}}{A_{Cu} } =\frac{d^{2}{Fe} }{ d^{2}{Cu} }$

$\frac{d^{2}{Fe} }{ d^{2}{Cu} } =\frac{p_{Fe} }{ p_{Cu} }$

$\frac{d{Fe} }{ d{Cu} } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}$

5 0

3 years ago

Many types of scientific equipment are used to perform different functions in the science lab.Which of the folllowing combinatio

faltersainse [42]

Answer:

Many types of scientific equipment are used to perform different functions in the science lab. Which of the following combinations of equipment would be needed to bring one liter of water to 85°C? a. ... Various pieces of safety equipment are used in the lab to provide protection against injury.

Explanation:

6 0

3 years ago