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3 years ago

14

Physics

2 answers:

anygoal [31]3 years ago

7 0

Hmmm hold on I’m gonna slove it and I’ll get back to you

labwork [276]3 years ago

3 0

Answer:

D. protons and neutrons

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Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o

Mila [183]

Answer:

a) $x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) x = 4.47 cm

c) $x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

$x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

$x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

$x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm$

c) The center of the mass of the beads realtive to the center of bead is:

$x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) $x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm$

6 0

3 years ago

Read 2 more answers

Ian walks 2km to his best friends house, the. Walks .5km to the library. Then makes a 2.5 km walk home. The entire walk took 3 h

max2010maxim [7]

1.66 km/h or 1 2/3 km/h

8 0

2 years ago

Read 2 more answers

A 1.0-kg block of aluminum is at a temperature of 50 Celsius. How much thermal energy will it lose when it’s temperature is redu

Ipatiy [6.2K]

Answer:

The lose of thermal energy is, Q = 22500 J

Explanation:

Given data,

The mass of aluminium block, m = 1.0 kg

The initial temperature of block, T = 50° C

The final temperature of the block, T' = 25° C

The change in temperature, ΔT = 50° C - 25° C

= 25° C

The specific heat capacity of aluminium, c = 900 J/kg°C

The formula for thermal energy,

= 1.0 x 900 x 25

= 22500 J

Hence, the lose of thermal energy is, Q = 22500 J

7 0

3 years ago

What is the period of a wave traveling 5 m/s if its wavelength is 20 m/s

Ilia_Sergeevich [38]

Speed of wave is given as

$v = 5 m/s$

Wavelength of the wave is given as

$\lambda = 20 m$

now from the formula of wave time period we can say

$speed = \frac{wavelength}{time period}$

$5 = \frac{20}{T}$

$T = \frac{20}{5}$

$T = 4 s$

so it will have time period of T = 4 s

7 0

3 years ago

With certain exceptions, Class E airspace extends upward from either 700 feet or 1,200 feet AGL to, but does not include,A) 14,5

DedPeter [7]

Answer:

B) 18,000 feet MSL

Explanation:

There are three-dimensional parts in the navigation airspace in the world. The class E airspace is mostly used in the regions with coastal areas that are relatively populated. If we consider certain forms of exceptions, the class E airspace can move in the upward direction to few feet (i.e. 1200 ft). However, this doesn't include 18,000 feet MSL.

6 0

3 years ago