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4 years ago

15

Explain to me an example of forces on an object you have used recently. Include all forces acting on the object and explain if t

he motion is balanced or unbalanced

1 answer:

masha68 [24]4 years ago

4 0

Answer:

a toy car being pushed by a kid, a force acts on the car

Explanation:

when the boy is pushing the car a force acts on the car which is the effort that is applied on the car and the motion is unbalanced because there will be constant stopping of the boy and also the force acting on the car

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In the United States, household electric power is provided at a frequency of 60 HzHz, so electromagnetic radiation at that frequ

grigory [225]

Answer:

the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Explanation:

Given the data in the question;

To determine the maximum intensity of an electromagnetic wave, we use the formula;

$I$ = $\frac{1}{2}$ ε₀cE $_{max$ ²

where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )

c is the speed of light ( 3 × 10⁸ m/s )

E $_{max$ is the maximum magnitude of the electric field

first we calculate the maximum magnitude of the electric field ( E $_{max$ )

E $_{max$ = 350/f kV/m

given that frequency of 60 Hz, we substitute

E $_{max$ = 350/60 kV/m

E $_{max$ = 5.83333 kV/m

E $_{max$ = 5.83333 kV/m × ( $\frac{1000 V/m}{1 kV/m}$ )

E $_{max$ = 5833.33 N/C

so we substitute all our values into the formula for intensity of an electromagnetic wave;

$I$ = $\frac{1}{2}$ ε₀cE $_{max$ ²

$I$ = $\frac{1}{2}$ × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²

$I$ = 45 × 10³ W/m²

$I$ = 45 × 10³ W/m² × ( $\frac{1 kW/m^2}{10^3W/m^2}$ )

$I$ = 45 kW/m²

Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

7 0

3 years ago

What role does a resistor play in an electrical circuit

kirza4 [7]

Answer: It opposes the flow of electrons.

Explanation: just did the quiz on

4 0

3 years ago

Read 2 more answers

I rent a small high pressure water sprayer to clean the outside of my house. The sprayer works like a super soaker with a hose b

viva [34]

Answer:

1. 80,000 Pa

2. 11.3 m/s

3. 12.5 m/s

Explanation:

<u>Question 1</u>

Pressure, $P=hg\rho$

Where h is the height that water is to reach, g is gravitational constant and $\rho$ is the density, in this case, we assume $\rho$ of pure water as $1000 Kg/m^3$

Assuming $g=10 m/s^{2}$

P=8*10*1000=80000 Pa

<u>Question 2</u>

Pressure can also be found by the formula

$P=0.5v^{2}\rho$ where v is the velocity

Equating the new formula of pressure to the formula used in question 1 above

$P=0.5v^{2}\rho=hg\rho$

Notice that $\rho$ is common hence

$0.5v^{2}=hg$

Making V the subject of the formula

$v^{2}=2hg$

$v=\sqrt 2hg$

In this case, h=8-1.6=6.4m and taking g as 10 m/s^{2}

$v=\sqrt 2*10*6.4=11.3137085 m/s$

Rounding off to 1 decimal place

v=11.3 m/s

<u>Question 3</u>

As already illustrated

$v=\sqrt 2hg$

Taking g as 9.8 and h now is 8m

$v=\sqrt 2*8*9.8$

v=12.52198067

Rounding off to 1 decimal place

v=12.5 m/s

6 0

3 years ago

A cylinder with a piston contains 0.200 mol of nitrogen at 1.50×105 Pa and 320 K . The nitrogen may be treated as an ideal gas.

Alja [10]

Answer:

$Q = -105 J$

Also we know that for cyclic process change in internal energy is always ZERO

Explanation:

First gas is compressed isobarically such that its volume is half of initial volume

So its temperature is also half

So heat given in this process is given as

$Q = nC_p \Delta T$

for diatomic gas we have

$C_p = \frac{7}{2} R$

so we will have

$Q = 0.200(\frac{7}{2}R)(160 - 320)$

$Q = -930.7 J$

Now in adiabatic process heat is not transferred

so in this process

$Q = 0$

so we have

$T_1V_1^{1.4-1} = T_2V_2^{1.4-1}$

$(160)(\frac{V}{2})^{0.4} = T_2(V)^{0.4}$

$T_2 = 121.26 K$

Now it is again reached to original pressure

so temperature will become initial temperature

so heat given in that part

$Q_3 = nC_v\Delta T$

here we know that

$C_v = \frac{5}{2}R$

$Q_3 = (0.200)(\frac{5}{2}R)(320 - 121.26)$

$Q_3 = 825.76 J$

So total heat given to the system is

$Q = -930.7 + 0 + 825.76$

$Q = -105 J$

Also we know that for cyclic process change in internal energy is always ZERO

3 0

3 years ago

Can sometimes solve problem 16

NISA [10]

It’s called the the magnetic field

7 0

3 years ago