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2 years ago

A ball is thrown vertically upward with an initial velocity

Physics

1 answer:

ivolga24 [154]2 years ago

5 0

Answer:

Explanation:

First we define our variables

V0=29.4

a=-9.8

V=0

We have to find the maximum displacement , which I will define as X

We use formula v^2=v0^2+2aX

All we do is substitute our values

0=29.4^2-19.6X

29.4^2=19.6X

X=29.4^2/19.6=44.1

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A factory has 1200 workers of which 720 are male and the rast are female what percent of workers are female

umka21 [38]

Answer:

Percent of Female Workers = 40%

Explanation:

The percentage of the female workers in the given group of workers can be easily found by the following formula:

$Percent\ of\ Female\ Workers = \frac{No.\ of\ Female\ Workers}{Total\ No.\ of\ Workers}\ x\ 100\%$

where,

Total No. of Workers = 1200

No. of Female Workers = Total Workers - No. of Male Workers

No. of Female Workers = 1200 - 720 = 480

Therefore,

$Percent\ of\ Female\ Workers = \frac{480}{1200}\ x\ 100\%$

<u>Percent of Female Workers = 40%</u>

7 0

3 years ago

Define investigation to show its scientific meaning.

Murrr4er [49]

Answer:

the action of investigating something or someone; formal or systematic examination or research.

Explanation:

This definition is provided by Oxford Languages

7 0

3 years ago

single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev

Goryan [66]

Answer:

The sound level of the 26 geese is $Z_{26}= 96.15 dB$

Explanation:

From the question we are told that

The sound level is $Z_1 = 81.0 \ dB$

The number of geese is $N = 26$

Generally the intensity level of sound is mathematically represented as

The intensity of sound level in dB for one goose is mathematically represented as

$Z_1 = 10 log [\frac{I}{I_O} ]$

Where I_o is the threshold level of intensity with value $I_o = 1*10^{-12} \ W/m^2$

$I$ is the intensity for one goose in $W/m^2$

For 26 geese the intensity would be

$I_{26} = 26 * I$

Then the intensity of 26 geese in dB is

$Z_{26} = 10 log[\frac{26 I }{I_o} ]$

$Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )$

$Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )$

From the law of logarithm we have that

$Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]$

$= 14.15 + 82$

$Z_{26}= 96.15 dB$

4 0

3 years ago

Calculate the weight of a new fast-food sandwich that has a mass of 0.1 kg (approximately the mass of a quarter pound). Think of

iragen [17]

Weight = (mass) x (gravity)

If you plan to sell these things on Earth, then the acceleration of gravity in the neighborhood of your drive-throughs will be 9.81 m/s².

Weight of each sandwich = (0.1 kg) x (9.81 m/s²).

Weight of each sandwich = 0.981 Newton.

This is only 1.9% less than 1 even Newton.

You should start by setting up one restaurant in New York, one in Chicago, one in LA, and maybe one in Miami or Tulsa. Sell it with a different name in each place, and see which name sells best.

You might want to try calling it

-- Isaac's burger

-- Gravity grub

-- Prism Patty

-- Mass 'o Meat

-- Unit-wich

and see if anything catches on.

I think I'd simply call it a "Newton Unit".

5 0

3 years ago

A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea

viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

$L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}$ (1)

time taken = 5 years

We know that :

time = $\frac{distance}{speed}$

5 = $\frac{L''}{v}$ (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

$\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

$\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

$t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}$

Solving the above eqn:

t'' = 20.56 years

4 0

3 years ago