PLS HELP!!!- The moon rotates around the Earth every 28 days. What is the frequency of the moon’s revolution?

What is the total distance, side to side, that the top of the building moves during such an oscillation? The New England Merchan

kramer

I think the question should be the below:

What is the total distance, side to side, that the top of the building moves during such an oscillation?

Answer is the below:

Acceleration .. a = (-) ω² x 
(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) 

x (max) = a(max) /ω² 

x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)

5 0

3 years ago

A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that

lina2011 [118]

Answer:

0.006075Joules

Explanation:

The final kinetic energy of the system is expressed as;

KE = 1/2(m1+m2)v²

m1 and m2 are the masses of the two bodies

v is the final velocity of the bodies after collision

get the final velocity using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

0.12(0.45) + 0/12(0) = (0.12+0.12)v

0.054 = 0.24v

v = 0.054/0.24

v = 0.225m/s

Get the final kinetic energy;

KE = 1/2(m1+m2)v

KE = 1/2(0.12+0.12)(0.225)²

KE = 1/2(0.24)(0.050625)

KE = 0.12*0.050625

KE = 0.006075Joules

Hence the final kinetic energy of the system is 0.006075Joules

5 0

3 years ago

For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mas

inn [45]

Answer:

Yes is large enough

Explanation:

We need to apply the second Newton's Law to find the solution.

We know that,

$F= ma$

And we know as well that

$a= \frac{v}{t}$

Replacing the aceleration value in the equation force we have,

$F= \frac{mv}{t}$

Substituting our values we have,

$F= \frac{(0.060)(55)}{4*10^{-3}}$

$F=825N$

The weight of the person is then,

$W = mg$

$W = (60)(9.8) = 558N$

We can conclude that force on the ball is large to lift the ball

6 0

3 years ago

51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00Ω by placing a 1.00-

Tanya [424]

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance $r= 5.00\ \Omega$

(a). We need to calculate the current

Using rule of loop

$E-IR-Ir=0$

$I=\dfrac{E}{R+r}$

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

$I=\dfrac{3.200}{1.00\times10^{3}+5.00}$

$I=3.184\times10^{-3}\ A$

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

$V=E-Ir$

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

$V=3.200-3.184\times10^{-3}\times5.00$

$V=3.18\ V$

(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf

$\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }$

$\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}$

Hence, This is the required solution.

5 0

3 years ago

A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o

Crank

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

$\delta = \frac{PL}{AE}$