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2 years ago

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An elevator accelerates upward at 1.2 m/s 2 . the acceleration of gravity is 9.8 m/s 2 . what is the upward force exerted by the

floor of the elevator on a(n) 62 kg passenger? answer in units of n.

1 answer:

Ira Lisetskai [31]2 years ago

6 0

Answer:

Explanation:

Fr= Fn-P

ma=Fn-mg

Fn= m(a+g)

Fn=62(1.2+9.8)

Fn= 682N

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It is known that the population mean for the verbal section of the SAT is 500 with a standard deviation of 100. In 2006, a sampl

kvasek [131]

Answer

given,

SAT is 500 with a standard deviation of 100.

a sample of 400 students whose family income was between $70,000 and $80,000 had an average verbal SAT score of 511.

sample mean = $\dfrac{standard\ deviation}{\sqrt{n}}$

= $\dfrac{100}{\sqrt{400}}$

= 5

95% confidence level is achieved within +/- 1.960 standard deviations.

1.960 standard deviations x 5 is equal to +/- 9.8

confidence interval = 511 - 9.8 --- 511 + 9.8

= 501.2-----520.8

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3 years ago

A carousel is moving with uniform circular motion. The centripetal acceleration of the ride is * 1 point circular

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> A pebble that is 3.81 m from the eye of a tornado has a tangential speed of 124 m/s. What is the magnitude of the pebble's centripetal acceleration?

Centripetal acceleration is calculated using the formula:

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a = 4,035.70 m/s^2 ~ 4036

> A dog sits 1.25 m from the center of a merry go round and revolves at a tangential speed of 2.0 m/s. If the dog's mass is 19.3 kg, what is the magnitude of the centripetal force on the dog?

a = v^2 / r

a = (2.0 m/s)^2 / 1.25 m

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> A ball attached to a string is whirled in a circle. If the string breaks, what causes the ball to move in a straight line?

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The orbital speed is related to distance from the formula:

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> What is the efficiency of a crane that requires 28,000 J of energy to lift a 200 kg object 10 meters?

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3 0

3 years ago

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

Fudgin [204]

Answer:

a) $speed=\pi cm/s$

b) $v_{max}=\frac{5\pi}{4} cm/s$

c) $a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

$y=Asin(\omega t)$

where:

A=amplitude

$\omega$ = angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

$\omega=\frac{2\pi}{T}$

so our angular frequency is:

$\omega=\frac{2\pi}{8s}$

$\omega=\frac{\pi}{4}$

so now we can build our equation:

$y=5sin(\frac{\pi}{4} t)$

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

$y=5sin(\frac{\pi}{4} t)$

$3=5sin(\frac{\pi}{4} t)$

so we solve for t:

$sin(\frac{\pi}{4} t)=\frac{3}{5}$

$\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})$

$t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})$

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

$y=5sin(\frac{\pi}{4} t)$

$y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}$

which simplifies to:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

and we can now substitute the t-value we found previously, so we get:

$y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))$

$y'=\pi$

so its velocity at that point is $\pi$ cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

$A=\frac{5\pi}{4}$

so:

$v_{max}=\frac{5\pi}{4} cm/s$

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

$y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)$

We can use the chain rule again to find this derivative so we get:

$y" =-\frac{5\pi}{4}sin(\frac{\pi}{4} t)*(\frac{pi}{4})$

so when simplified we get:

$y"=-\frac{5\pi^{2}}{16}sin(\frac{\pi}{4} t)$

its amplitude is:

$A=\frac{5\pi^{2}}{16}$

so its maximum acceleration is:

$a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

7 0

3 years ago

A summary of a source is usually

Mekhanik [1.2K]

Answer:

A shorter than the original source and in the researcher's words

Explanation:

The summary is an abridged version of the original source and in the researcher's own very words.

Summaries gives an over-arching perspective and excludes implicit details from a given text.

A summary should not be detailed and must avoid overt illustrations. They must capture the true essence of piece leaving out flowery details.

A summary should be lesser in length than the original piece. Any third party reader should immediately be able to grab the details of the original piece from the summary.

5 0

3 years ago

Read 2 more answers

A decorative plastic film on a copper sphere of 10 mm diameter in an oven at 750C. Upon removal from the oven, the sphere is sub

AlladinOne [14]

Answer:

3.26 secs

Explanation:

Diameter of sphere ( D )= 10 mm

T1 = 75°C

P = 1 atm

T∞ = 23°C

T2 = 35°c

Velocity = 10 m/s

<u>Determine how long it will take to cool the sphere to 35°C</u>

<em>Using the properties of copper and air given in the question</em>

Nu = 2 + (Re)^0.8 (Pr)^0.33

hd / k = 2 + ( vd/v )^0.8 (Pr)^0.33

∴ h ≈ 2594.7 W/m^2k

Given that :

(T2 - T∞) / ( T1 - T∞ ) = exp [ ( -hA / pv CP ) t ]

( 35 - 23 ) / ( 75 - 23 ) = exp [ - 2594.7 * 6 * t / 8933 * 387 * 10 * 10^-3 ]

= ln ( 12/52 ) = -1.466337069 = - 0.45032919 * t

∴ t ≈ 3.26 secs ( -1.466337069 / -0.45032919 )

6 0

3 years ago