PHYSICS COLLEGE

A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle with a constant angular speed of 120 rev/min. The cross-sectional area of thewire is 0.014 cm2. Calculate the elongation of the wire when the mass is a) at its lowest point of the path and b) at the highest point of its path

Answers

Answer 1

Answer:

Answer:

a) the elongation of the wire when the mass is at its lowest point on the path = 0.5 cm

b) the elongation of the wire when the mass is at its highest point on the path = 0.42 cm

Explanation:

Given that;

the angular speed $\omega = 120 \ rev/min$

Then converting it to rad/s ; we have:

= $(120 \ rev/min )((2 \ \pi \ rad )/(1 \ rev) ) ((1 \ min )/(60 \ s) )$

= 12.57 rad/s

The cross-sectional area of the wire A = 0.014 cm²

A = (0.014 cm²) ( $(10^(-4) \ m^2)/(1 \ cm^3)$ )

A = $0.014*10^(-4) \ m^2$

mass (m) = 12.0 kg

R = 0.5 m

g = 9.8 m/s²

To calculate for the mass when its at the lowest point of the path; we use the Newton's second law of motion; which is expressed as:

$T - mg = ma_(rad)$

where;

$a_rad = ( radical \ acceleration ) = R \omega^2$

Now; we can rewrite our equation as;

$T -mg = m R \omega ^2$

$T = mR \omega^2 + mg$

$T = m( R \omega^2 + g)$

Replacing our given values ; we have:

$T = 12.0( 0.5(12.57)^2 + 9.8)$

$T = 12.0( 0.5(158.0049) + 9.8)$

$T = 12.0( 79.00245 + 9.8)$

$T = 12.0( 88.80245)$

T = 1065.6294 N

T ≅ 1066 N

Determining the elongation $\delta l$ in the wire by using the equation

Y = $(Tl)/(AY)$

Making $\delta l$ the subject of the formula; we have

$\delta l = (Tl)/(AY)$

where ;

l = length of the wire

T =Tension in the wire

A = cross - sectional area

Y = young's modulus

Then;

$\delta l = ((1066 N)(0.5m))/((0.014*10^(-4)m^2)(7.0*10^(10)Pa))$

$\delta l$ = $( 0.00544 m ) *((10 ^2 cm)/(1m) )$

$\delta l$ = 0.5 cm

Thus, the elongation of the wire when the mass is at its lowest point on the path = 0.5 cm

Using Newton's second law of motion also for the mass at its highest point of the path; we have:

$T +mg = m R \omega ^2$

$T = mR \omega^2- mg$

$T = m( R \omega^2 - g)$

Replacing our given values ; we have:

$T = 12.0( 0.5(12.57)^2 - 9.8)$

$T = 12.0( 0.5(158.0049)-9.8)$

$T = 12.0( 79.00245 - 9.8)$

$T = 12.0( 69.20245)$

T = 830.4294 N

T = 830 N

Determining the elongation $\delta l$ in the wire by using the equation

Y = $(Tl)/(AY)$

Making $\delta l$ the subject of the formula; we have

$\delta l = (Tl)/(AY)$

where ;

l = length of the wire

T =Tension in the wire

A = cross - sectional area

Y = young's modulus

Then;

$\delta l = ((830 N)(0.5m))/((0.014*10^(-4)m^2)(7.0*10^(10)Pa))$

$\delta l$ = $( 0.00424 m ) *((10 ^2 cm)/(1m) )$

$\delta l$ = 0.42 cm

Thus, the elongation of the wire when the mass is at its highest point on the path = 0.42 cm

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A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.

Answers

Answer:

4.93 m

Explanation:

According to the question, the computation of the height is shown below:

But before that first we need to find out the speed which is shown below:

As we know that

$Speed = (Distance)/(Time)$

$Speed = (5)/(0.504)$

= 9.92 m/s

Now

$v^2 - u^2 = 2* g* h$

$9.92^2 = 2* 9.98 * h$

98.4064 = 19.96 × height

So, the height is 4.93 m

We simply applied the above formulas so that the height i.e H could arrive

Final answer:

The height of the water slide is 5.04 meters.

Explanation:

The problem described in this question involves a water slide, where swimmers start from rest at the top and leave the slide traveling horizontally. To determine the height of the slide, we can use the equations of motion in the horizontal direction. The horizontal displacement (x) is given as 5.00 m and the time (t) is given as 0.504 s. Assuming no friction or air resistance, we can use the equation x = v*t, where v is the horizontal velocity. Rearranging the equation, we can solve for v, which is equal to x/t. Substituting the given values, we have v = 5.00 m / 0.504 s = 9.92 m/s. The horizontal velocity (v) is constant throughout the motion, so we can use the equation v = sqrt(2*g*H), where g is the acceleration due to gravity (9.8 m/s^2) and H is the height of the slide. Rearranging the equation, we can solve for H, which is equal to v^2 / (2*g). Substituting the known values, we have H = (9.92 m/s)^2 / (2*9.8 m/s^2) = 5.04 m.

PLEASE HELP DUE BEFORE 11:30 TODAY!!!!Which of the following quantities is NOT a vector quantity?
A. 926 m to the north
B. 5.2 m/s to the west
C. 46 m down
D. 12.3 m/s faster

Answers

Answer:

D is not the a vector quantities

A helium balloon ride lifts up passengers in a basket. Assume the density of air is 1.28 kg/m3 and the density of helium in the balloon is 0.18 kg/m3. The radius of the balloon (when filled) is R = 4.9 m. The total mass of the empty balloon and basket is mb = 121 kg and the total volume is Vb = 0.073 m3. Assume the average person that gets into the balloon has a mass mp = 73 kg and volume Vp = 0.077 m3. 1)What is the volume of helium in the balloon when fully inflated? m3 2)What is the magnitude of the force of gravity on the entire system (but with no people)? Include the mass of the balloon, basket, and helium. N

Answers

Answer:

1) The volume of helium in the ballon when is fully inflated is 492.8070 m³

2) The magnitude of the force of gravity (with no people) is 869.3119 N

Explanation:

Given:

ρair = density of air = 1.28 kg/m³

ρhelium = density of helium = 0.18 kg/m³

R = radius of balloon = 4.9 m

mtotal = 121 kg

Vtotal = 0.073 m³

mp = average mass per person = 73 kg

Vp = 0.077 m³

g = gravity = 9.8 m/s²

Questions:

1) What is the volume of helium in the balloon when fully inflated, Vhelium = ?

2) What is the magnitude of the force of gravity on the entire system (but with no people), Fg = ?

1) The volume of helium in the ballon when is fully inflated

$V_(helium) =(4)/(3) \pi R^(3) =(4)/(3) \pi *4.9^(3) =492.8070m^(3)$

2) First, you need to calculate the mass of helium

$m_(helium) =\rho _(helium) *V_(helium) =0.18*492.8070=88.7053kg$

The magnitude of the force of gravity (with no people)

$F_(g) =m_(helium) *g=88.7053*9.8=869.3119N$

A revolutionary war cannon, with a mass of 2260 kg, fires a 21 kg ball horizontally. The cannonball has a speed of 105 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?

Answers

Answer:

0.97566 m/s

Explanation:

$m_1$ = Mass of cannon = 2260 kg

$v_1$ = Velocity of cannon

$m_2$ = Mass of ball = 21 kg

$v_2$ = Velocity of ball = 105 m/s

As the momentum of the system is conserved we have

$m_1v_1=m_2v_2\n\Rightarrow v_1=(21* 105)/(2260)\n\Rightarrow v_1=0.97566\ m/s$

The velocity of the cannon is 0.97566 m/s

A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The bullet gets embedded in the block. The speed of the block immediately after the collision is:________. A) cannot be found because we don't know whether the surface is frictionless.
B) is 0.21 km/s.
C) is 65 m/s.
D) is 9.3 m/s.
E) None of these is correct

Answers

Answer:

4.25m/s

E. None of the option is correct

Explanation:

Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.

Mathematically.

mu + MU = (m+M)v

m and M are the masses of the bullet and the block respectively

u and U are their respective velocities

v is their common velocity

from the question, the following parameters are given;

m = 20g = 0.02kg

u = 960m/s

M = 4.5kg

U =0m/s (block is at rest)

Substituting this values into the formula above to get v;

0.02(960)+4.5(0) = (0.02+4.5)v

19.2+0 = 4.52v

4.52v = 19.2

Dividing both sides by 4.52

4.52v/4.52 = 19.2/4.52

v = 4.25m/s

Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s

An object essentially at infinity is moved to a distance of 90 cm in front of a thin positive lens. In the process its image distance triples. Determine the focal length of the lens.

Answers

Answer:

67.5 cm

Explanation:

u = - 90 cm, v = 3 x u = 3 x 90 = 270 cm

let f be the focal length

Use lens equation

1 / f = 1 / v - 1 / u

1 / f = 1 / 270 + 1 / 90

1 / f = 4 / 270

f = 67.5 cm

Final answer:

To determine the focal length of the lens, we use the lens formula and set up an equation based on the given information. Solving for the image distance, we find that it is zero, indicating the image is formed at infinity. Therefore, the focal length of the lens is 90 cm.

Explanation:

To determine the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u

Where f is the focal length, v is the image distance, and u is the object distance.

Given that the image distance triples when the object is moved from infinity to 90 cm in front of the lens, we can set up the following equation:

1/f = 1/(3v) - 1/(90)

Multiplying through by 90*3v, we get:

90*3v/f = 270v - 90*3v

90*3v/f = 270v - 270v

90*3v/f = 0

Simplifying further, we find that: v = 0

When the image distance is zero, it means the image is formed at infinity, so the lens is focused at the focal point. Therefore, the focal length of the lens is 90 cm.

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A quarterback claims that he can throw the football a horizontal distance of 167 m. Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 33.1 ° above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional.

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A electromagnetic wave of light has a wavelength of 500 nm. What is the energy (in Joules) of the photon representing the particle interpretation of this light?