What is The most basic unit of an Element on the Periodic Table

The mini-refrigerator fire was most likely caused by what type of wiring?

Nimfa-mama [501]

Answer: The copper wiring most likely caused the mini-refrigerator fire.

Explanation: A mini-refrigerator may have one or more evaporator fan motors. These motors have copper wiring. In some situation these copper wires gets overheated. It may result to the firing of the mini-refrigerator.

Therefore, the mini-refrigerator fire was most likely caused by copper wiring.

7 0

3 years ago

An elevator motor provides 45.0 kW of power while lifting an elevator 35.0 m. If the elevator contains seven passengers each wit

mylen [45]

Find how much work ∆W is done by the motor in lifting the elevator:

P = ∆W / ∆t

where

• P = 45.0 kW = power provided by the motor

• ∆W = work done

• ∆t = 20.0 s = duration of time

Solve for ∆W :

∆W = P ∆t = (45.0 kW) (20.0 s) = 900 kJ

In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is M = (m + 490.0) kg, where m is the mass of the elevator alone. Then

∆W = M g h

where

• g = 9.80 m/s² = acceleration due to gravity

• h = 35.0 m = distance covered by the elevator

Solve for M, then for m :

M = ∆W / (g h) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg

m = M - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg

4 0

3 years ago

Blythe and Geoff are ice-skating together. Blythe has a mass of 40 kg and Geoff has a mass of 79 kg. Blythe pushes Geoff in the

Deffense [45]

9.9 m/s

calculations below

4 0

3 years ago

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A box of mass m1 rests on a smooth, horizontal floor next to a box of mass m2. Suppose the force of 20.0 N pushes on two boxes o

meriva

Answer:

Explanation:

Given

acceleration of system a =1.2 m/s^2

Normal Force N=4.45 N

Force exerted F=20 N

Thus

$F=(m_1+m_2)a$

$\frac{20}{1.2}=m_1+m_2$

$16.67=m_1+m_2$ -------1

Normal reaction $N=m_2a$

$4.45=m_2\times 1.2$

$m_2=3.70 kg$

therefore $m_1=16.67-3.70$

$m_1=12.96 kg$

8 0

3 years ago

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The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t

Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

$PE=\frac{GMm}{r}$

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

$m = \rho V$

$m = (1030Kg/m^3)(7*10^8m^3)$

$m = 7.210*10^{11}Kg$

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

$r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m$

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

$r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m$

PART A) Potential energy when the ocean is at its furthest point to the moon,

$PE_1 = \frac{GMm}{r_1}$

$PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}$

$PE_1 = 9.05*10^{15}J$

PART B) Potential energy when the ocean is at its closest point to the moon

$PE_2 = \frac{GMm}{r_2}$

$PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}$

$PE_2 = 9.361*10^{15}J$

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

$\Delta KE = \Delta PE$

$\frac{1}{2}mv^2 = PE_2-PE_1$

$v=\sqrt{2(PE_2-PE_1)/m}$

$v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}$

$v = 29.4m/s$

8 0

3 years ago