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4 years ago

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A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 12-m-long cable. You are (un

wisely!) standing directly beneath the point from which the wrecking ball is hung when you notice that the ball has just been released and is swinging directly toward you. How much time do you have to move out of the way?

2 answers:

leva [86]4 years ago

8 0

Because you know that gravity is in m/s^2 so, period will be measured in seconds. You know the cable is 12m long and gravity is 9.81 solve for T (period) 2π12sqrt(9.81)=6.94922

lisov135 [29]4 years ago

8 0

Answer:

Time to move out of the way = 1.74 s

Explanation:

This case is similar to simple pendulum. The period of simple pendulum is given by $2\pi \sqrt{\frac{l}{g} }$ .

Period of wrecking ball $=2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{12}{9.81}}=6.95s$

Time to move out of the way is one fourth of period = 6.95/4 = 1.74 seconds.

Time to move out of the way = 1.74 s

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In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le

expeople1 [14]

Answer:

The value is $m \approx 310$

Explanation:

From the question we are told that

The focal length of the objective is $f_o = 1.0 \ cm$

The focal length of the eyepiece is $f_e = 2.0 \ cm$

The tube length is $L = 25 \ cm$

Generally the magnitude of the overall magnification is mathematically represented as

$m = m_o * m_e$

Where $m_o$ is the objective magnification which is mathematically represented as

$m_o = \frac{L}{f_o }$

=> $m_o = \frac{25}{1 }$

=> $m_o = 25$

$m_e$ is the eyepiece magnification which is mathematically evaluated as

$m_e = \frac{L }{f_e }$

$m_e = \frac{25 }{ 2}$

$m_e = 12.5 \ cm$

So

$m = 25 * 12.5$

$m \approx 310$

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3 years ago

A mass of 2000 kg is raised 5.0 m in 10 seconds what is the potential energy of the mass at this height?

kvv77 [185]

Answer:

A. 98,000 J

Explanation:

The gravitational potential energy of an object is given by

U = mgh

where

m is the mass of the object

g is the gravitational acceleration

h is the heigth above the ground

In this problem,

m = 2000 kg

g = 9.8 m/s^2

h = 5.0 m

Substituting into the equation, we find

$U=(2000 kg)(9.8 m/s^2)(5.0 m)=98,000 J$

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4 years ago

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 43.0 m/s; when it leav

kotykmax [81]

Answer:

$-7.352\times 10^3$ N

Explanation:

We are given that

Mass of baseball=m=0.145 kg

Initial velocity of ball=u=43 m/s

$\theta=31^{\circ}$

Speed, v= $51 m/s$

Time, t=1.71 ms= $1.71\times 10^{-3} s$

$1 ms=1.71\times 10^{-3} s$

Horizontal component of the average force on the ball= $\frac{m(vcos\theta-u)}{t}$

Horizontal component of the average force on the ball $=\frac{0.145(-51cos31-43)}{1.71\times 10^{-3}}$

Horizontal component of the average force on the ball= $-7.352\times 10^3$ N

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4 years ago

Problem 25.40 What is the energy (in eV) of a photon of visible light that has a wavelength of 500 nm

Lapatulllka [165]

Answer:

E = 2.48 eV

Explanation:

The energy of a photon is given by the following formula:

E = hυ

where,

E = Energy of Photon = ?

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υ = frequency of photon = c/λ

Therefore,

E = hc/λ

where,

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light = 500 nm = 5 x 10⁻⁷ m

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5 x 10⁻⁷ m)

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<u>E = 2.48 eV</u>

3 0

4 years ago

How much pressure is applied to the

den301095 [7]

Answer:

306.12N/m^2

Explanation:

Mass = 30 kg

g = 9.8m/s2

F = mg

F = 30 x 9.8

F = 294N

L = 0.98m

Since the stilts is square, then the area is L^2

A = L^2

A = 0.98^2

A = 0.9604m^2

P =?

P = F/A

P = 294/0.9604

P = 306.12N/m^2

The pressure exerted by the man is 306.12N/m^2

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3 years ago