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zzz [600]
3 years ago
8

A 55.0-kg lead ball is dropped from the Leaning Tower of Pisa. The tower is 55.0 m high. What is the speed of the ball after it

has traveled 4.20 m downward
Physics
1 answer:
PolarNik [594]3 years ago
4 0

Answer:

The speed of the ball is 9.07 m/s.

Explanation:

Given that,

Mass of the lead ball, m = 55 kg

Height of the tower, h = 55 m

We need to find the speed of the ball it has traveled 4.20 m downward, x = 4.2 meters

The initial speed of the ball will be 0 as it was at rest initially. Let v is the speed of the ball after it has traveled 4.20 m downward. It is a case of equation of motion such that :

v^2-u^2=2ax

v^2=2ax

Here, a = g

v^2=2\times 9.8\times 4.2

v = 9.07 m/s

So, the speed of the ball is 9.07 m/s. Therefore, this is the required solution of given condition.

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The slider of mass m is released from rest in position A and slides without friction along the vertical-plane guide shown. Deter
Anuta_ua [19.1K]

The value of normal force as the slider passes point B is

  • 6 mg

The value of h when the normal force is zero

  • 3R/2

<h3>How to solve for the normal force</h3>

The normal force is calculated using the work energy principle which is applied as below

K₁ + U₁ = K₂

k represents kinetic energy

U represents potential energy

the subscripts 1,2 , and 3 = a, b, and c

for 1 to 2

K₁ + W₁ = K₂

0 + mg(h + R) = 0.5mv²₂

g(h + R) = 0.5v²₂

v²₂ = 2g(1.5R + R)

v²₂ = 2g(2.5R)

v²₂ = 5gR

Using summation of forces at B

Normal force, N  = ma + mg

N = m(a + g)

N = m(v²₂/R + g)

N = m(5gR/R + g)

N = 6mg

for 1 to 3

K₁ + W₁ = K₃ + W₃

0 + mgh = 0.5mv²₃ + mgR

gh = 0.5v²₃ + gR

0.5v²₃ = gh - gR

v²₃ = 2g(h - R)

at C

for normal force to be zero

ma = mg

v²₃/R = g

v²₃ = gR

and v²₃ = 2g(h - R)

gR = 2gh - 2gR

gR + 2gR = 2gh

3gR = 2gh

3R/2 = h

Learn more about normal force at:

brainly.com/question/20432136

#SPJ1

8 0
1 year ago
What is special about nitrogen, and what is its main function in the atmosphere
Vlad [161]
Nitrogen is the most abundant of the gases present in the atmosphere. 78 percent of the atmospheric air comprises of nitrogen, oxygen makes up for 21 percent and all other gases make up for the remaining one percent. Oxygen is a highly flammable gas and in the absence of nitrogen it would not have been possible to utilize this atmospheric oxygen, hence the presence of nitrogen reduces its flammability and also neutralizes the toxicity of other gases.
5 0
3 years ago
a) What is the average useful power output of a person who does6.00×10^6Jof useful work in 8.00 h? (b) Working at this rate, how
NARA [144]

Answer:

208.33 W

141.26626 seconds

Explanation:

E = Energy = 6\times 10^6\ J

t = Time taken = 8 h

m = Mass = 2000 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height of platform = 1.5 m

Power is obtained when we divide energy by time

P=\frac{E}{t}\\\Rightarrow P=\frac{6\times 10^6}{8\times 60\times 60}\\\Rightarrow P=208.33\ W

The average useful power output of the person is 208.33 W

The energy in the next part would be the potential energy

The time taken would be

t=\frac{E}{P}\\\Rightarrow t=\frac{mgh}{208.33}\\\Rightarrow t=\frac{2000\times 9.81\times 1.5}{208.33}\\\Rightarrow t=141.26626\ s

The time taken to lift the load is 141.26626 seconds

5 0
3 years ago
A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightl
USPshnik [31]

Given Information:  

Wavelength =  λ = 39.1 cm = 0.391 m

speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

tension = T = 160 N

Required Information:

Length of the vibrating string = L = ?

Answer:

Length of the vibrating string = 0.28 m

Explanation:

The frequency of beautiful note is

f = v/λ

f = 344/0.391

f = 879.79 Hz

As we know, the speed of the wave is

v = √T/μ

v = √160/0.00066

v = 492.36 m/s

The wavelength of the string is

λ = v/f

λ = 492.36/879.79

λ = 0.5596 m

and finally the length of the vibrating string is

λ = 2L

L = λ/2

L = 0.5596/2

L = 0.28 m

Therefore, the vibrating section of the violin string is 0.28 m long.

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