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3 years ago

9

Curious George is whirling a 2.0 kg bunch of bananas on a smooth floor in a circular path having a radius of 0.50 m. What force

must he apply to keep the motion constant so that the bananas complete one revolution every 4.0 seconds

1 answer:

Sedaia [141]3 years ago

4 0

Answer:

The force will be $F_{c}=2.47 N$

Explanation:

Let's use the centripetal force equation.

$F_{c}=m\omega ^{2}R$

Where:

m is the mass of the bunch of bananas

ω is the angular speed

R is the radius

Now, 1 rev every 4 seconds or 0.25 rev/sec is the angular speed, but we need to write this speed in rad per second.

$\omega =0.25\frac{rev}{s}=0.25*2\pi \frac{rad}{s}=1.57 \frac{rad}{s}$

FInally, the force will be:

$F_{c}=2.0*1.57^{2}*0.5$

$F_{c}=2.0*1.57^{2}*0.5$

$F_{c}=2.47 N$

I hope it helps you!

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Irina finds an unlabeled box of fine needles, and wants to determine how thick they are. A standard ruler will not do the job, a

Vsevolod [243]

Answer:

$d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m$

Explanation:

The expression which represent the first diffraction minima by a circular aperture is given by $d sin\Theta =1.22\lambda$ --------eqn 1

The angle through which the first minima is diffracted is given by $tan\Theta =\frac{y_1}{D}$ ---------eqn 2

As $\Theta$ is very small so we can write $sin\Theta =tan\Theta$

So from eqn 1 and eqn 2 we can write

$y_1=\frac{1.22\lambda D}{d}$ --------eqn 3

Here $y_1$ is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture

It is given that diameter of circular aperture is 14.7 cm so $y_1=\frac{14.7}{2}=7.35 \ cm$

Now putting all these value in eqn 3

$d=\frac{1.22\lambda D}{y_1}$

$d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m$

8 0

3 years ago

Water is entering the prism at a rate of A m^3/hr. The prism is empty at time 0. Express the depth d of the water in meters in t

zavuch27 [327]

This question is incomplete, the complete question is;

The picture shows a triangular prism. The end of prism are equilateral triangles with x meters. the other dimension of the prism is L meters

a) Find the volume V in terms of x and L

b) Water is entering the prism at a rate of A m³/hr. The prism is empty at time 0. Express the depth d of the water in meters in terms of A, the length of time t the water has been entering the trough, and the length L of the prism.

Answer:

a) the volume V in terms of x and L is ((√3/4)x²L) m³

b) required expression is (2/(3)^(1/u))√(At/L)

Explanation:

Given that;

form the question and image below;

triangular prism ends are equilateral triangle

side length = x meter

Dimension of the prism = L meter

Area of the equilateral triangle = √3/4 (side)² = √3/4 (x)² meter

Volume of the triangular prism = Area × height

= √3/4 (x)² × L

V = ((√3/4)x²L) m³

Therefore, the volume V in terms of x and L is ((√3/4)x²L) m³

b)

Rate of water entering = A m³/hr

Depth of water tank = d meter

Time = t

Length of prism = L

now Rate of water entering is A m³/hr

dv/d = A [ V = ((√3/4)x²L) m³ ]

and

dv/dt = √3/4 [2x dx/dt ] L { L is constant }

so

A = √3/4 [2x dx/dt ] L

∫A dt = √3/2 [ Lx dx ] { Integrate both sides}

At = √3/2 × Lx × x²/2

x² = uAt / √3L { we find square root of both sides}

x = √( uAt / √3L )

x = (2/(3)^(1/u))√(At/L)

Therefore; required expression is (2/(3)^(1/u))√(At/L)

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2 years ago

Identify the elements that come before iron (Fe) in the periodic table Fe, Co, Cu, K, Ni, Mn

harkovskaia [24]

FE is iron CO is cobalt CU is copper K is potassium NI is nickle MN is magnemese

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3 years ago

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Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration

eimsori [14]

The complete text of the problem is:

<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>

<em />

<u>Solution:</u>

1) Acceleration: $-2336 m/s^2$ on the hardwood floor, $-382 m/s^2$ on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

$v^2 - u^2 = 2ad$

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

a = g = 9.8 m/s^2 is the acceleration of gravity

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

$v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s$

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

$d = 1.8 mm = 0.0018 m$

So we can find the acceleration by using again the equation

$v'^2 - v^2 = 2ad$

Solving for a,

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2$

For the carpeted floor instead,

$d=1.1 cm = 0.011 m$

therefore the acceleration is

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2$

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

$a=\frac{v'-v}{t}$

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

$a=-2336 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms$

For the carpeted floor,

$a=-382 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms$

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

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3 years ago

What is the momentum of a 50-kilogram ice skater gliding across the ice at a speed of 5 m/s? (1 point)

serious [3.7K]

Momentum is a term used to quantify the motion of an object has. It is calculated as the the product of the object's mass and the velocity. It is expressed as:

Momentum = m x v
Momentum = 50 kg x 5 m/s
Momentum = 250 kg m/s

Therefore, the correct answer is the last option.

3 0

3 years ago

Read 2 more answers