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3 years ago

What is the acceleration aaa of the refrigerator 4 ss after the person begins pushing on it with a force of 400 NN

Physics

1 answer:

ruslelena [56]3 years ago

3 0

Answer:

2 ms^2

Explanation:

Given that the mass of the refrigerator is 200Kg

The net force acting on the refrigerator is 400N

The time takes is 4s

From Newtons's second law of motion;

F =ma

a = F/m

a = 400/200

a = 2 ms^2

This is a constant acceleration

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Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.

olga2289 [7]

Answer:

$NO_{1.499}$

Explanation:

Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:

$n_{N} = \frac{36.86\,kg}{14.006\,\frac{kg}{kmol} }$

$n_{N} = 2.632\,kmol$

$n_{O} = \frac{63.14\,kg}{15.999\,\frac{kg}{kmol} }$

$n_{O} = 3.946\,kmol$

Ratio of kilomoles oxygen to kilomole nitrogen is:

$n^{*} = \frac{3.946\,kmol}{2.632\,kmol}$

$n^{*}= 1.499$

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

$NO_{1.499}$

8 0

4 years ago

Read 2 more answers

Voltage needed to raise current to 3.75a using 20,20,200 resistor set

Varvara68 [4.7K]

<u>Answer:</u> The voltage needed is 35.7 V

<u>Explanation:</u>

Assuming that the resistors are arranged in parallel combination.

For the resistors arranged in parallel combination:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

We are given:

$R_1=20\Omega\\R_2=20\Omega\\R_3=200\Omega$

Using above equation, we get:

$\frac{1}{R}=\frac{1}{20}+\frac{1}{20}+\frac{1}{200}\\\\\frac{1}{R}=\frac{10+10+1}{200}\\\\R=\frac{200}{21}=9.52\Omega$

Calculating the voltage by using Ohm's law:

$V=IR$ .....(1)

where,

V = voltage applied

I = Current = 3.75 A

R = Resistance = $9.52\Omega$

Putting values in equation 1, we get:

$V=3.75\times 9.52\\\\V=35.7V$

Hence, the voltage needed is 35.7 V

7 0

3 years ago

Read 2 more answers

Is it true of false that the slope of a line equals the average speed of the object.

kupik [55]

IF the object's motion is graphed, and IF time is plotted
along the x-axis and the object's position is plotted along
the y-axis, then the slope of the graph at every point is the
object's speed at that instant of time.

8 0

3 years ago

A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat

Triss [41]

Answer:

2.1406 × $10^6$ m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

$qV=\frac{1}{2}mv^2$

here V is the potential difference

we know that mass of proton = 1.67× $10^{-27}$ kg

we have given speed =50000m/sec

so potential difference $V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045$

now mass of electron =9.11× $10^{-31}$

so for electron

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec$

so the velocity of electron will be 2.1406× $10^6$ m/sec

4 0

3 years ago

The rear window in a car is approximately a rectangle, 1.3 m wide and 0.30 m high. The inside rear-view mirror is 0.50 m from th

maria [59]

Answer:

Height of mirror 0.075 m

width of mirror 0.325 m

Explanation:

given data

wide = 1.3 m

high = 0.30 m

driver’s eyes = 0.50 m

rear window = 1.50 m

solution

we take here height / width of the mirror is

height / width = $\frac{h}{w}$ .................1

and

height /width of the window is

height /width = $\frac{h_w}{w_w}$ .................2

and

distance of eye / window by the mirror is

distance of eye / window = $\frac{x_e}{x_w}$ .................3

so here

θ = θi = θr ....................4

and tanθ for vertical is

tanθ = $\frac{h}{x_e}$

tanθ = $\frac{h_w}{(x_e + x_w)}$ ....................5

h = $h_w \times \frac{x_e}{(x_e + x_e)}$ ....................6

put here value and we get

$h = 0.30 \times \frac{0.50}{(0.50 + 1.50)}$

h = 0.075 m

and

when we take here tanθ for horizontal than it will be

tanθ = $\frac{w}{x_e}$

tanθ = $\frac{w_w}{(x_e + x_w)}$ .......................7

$w = w_w \times \frac{x_e}{(x_e + x_w)}$ ....................8

put here value and we get

$w = 1.3 \times \frac{0.50}{(0.50 + 1.50)}$

w = 0.325 m

7 0

3 years ago