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3 years ago

State Newton's first law of motion .

Physics

1 answer:

malfutka [58]3 years ago

4 0

Answer:

dis mans newton says if a body is at rest or moving at a constant speed in a straight linne it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force

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A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

Ugo [173]

Answer:

$f_{e}$ = 1.9 cm

Explanation:

The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective

M = M₀ $m_{e}$

Where M₀ is the magnification of the objective and $m_{e}$ is the magnification of the eyepiece.

The eyepiece is focused to the near vision point (d = 25 cm)

$m_{e}$ = 25 / $f_{e}$

The objective is focused on the distances of the tube (L)

M₀ = -L / f₀

Substituting

M = - L/f₀ 25/ $f_{e}$

1) Let's look for the focal length of the eyepiece (faith)

$f_{e}$ = - L 25 / f₀ M

M = 400X = -400

$f_{e}$ = - 12 25 /0.40 (-400)

$f_{e}$ = 1.875 cm

Let's approximate two significant figures

$f_{e}$ = 1.9 cm

8 0

3 years ago

Describe the difference between particle motion in solids and liquids.

4vir4ik [10]

Answer:

Particles in a: gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

i hope this helps your answer

5 0

3 years ago

Read 2 more answers

Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the

Scrat [10]

Answer: $1.8\°$

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d=3.4(10)^{-5}m$ is the width of the slit

$\lambda=540 nm=540(10)^{-9}m$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the second-order diffraction angle is given when $n=2$ , hence equation (1) becomes:

$dsin\theta_{2}=2\lambda$ (2)

Now we have to find the value of $\theta_{2}$ :

$sin\theta_{2}=\frac{2\lambda}{d}$ (3)

Then:

$\theta_{2}=arcsin(\frac{2\lambda}{d})$ (4)

$\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})$ (5)

Finally:

$\theta_{2}=1.8\°$ (6)

5 0

3 years ago

Below is a circuit schematic of sources and resistors (Figure 3). VS = 10V , R1 = 100Ω, R2 = 50Ω, R3 = 25Ω, IS = 2A. Calculate t

lorasvet [3.4K]

Answer:

$V_3\approx 4.28\,\,V$

$I_1=0.0572\,\,amps$

$I_3\approx 0.171\,\,amps$

Explanation:

Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.

So we first find the equivalent resistance for the two resistors in parallel:

$\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega$

By knowing this, we can estimate the total current through the circuit,:

$Vs=I\,*\,(\frac{100}{3} +25)\\10=I\,*\,\frac{175}{3} \\I=\frac{30}{175} \,amps$

So approximately 0.17 amps

and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

$V_3=\frac{30}{175} *\,25=\frac{30}{7} \approx 4.28\,\,V$

So now we know that the potential drop across the parellel resistors must be:

10 V - 4.28 V = 5.72 V

and with this info, we can calculate the current through R1 using Ohm's Law:

$I_1=\frac{V_1}{R_1} =\frac{5.72}{100} =0.0572\,\,amps$

4 0

3 years ago

The mass of a hot-air balloon and its cargo (not including the air inside) is 170 kg. The air outside is at 10.0°C and 101 kPa.

scoundrel [369]

Answer:

108.37°C

Explanation:

P₁ = Initial pressure = 101 kPa

V₁ = Initial volume = 530 m³

T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K

P₂ = Final pressure = 101 kPa (because it is open to atmosphere)

V₂ = Final volume = 530 m³

P₁V₁ = n₁RT₁

⇒101×530 = n₁RT₁

⇒53530 J = n₁RT₁

P₂V₂ = n₂RT₂

⇒53530 J = n₂RT₂

$\frac{m_1}{m_2}=\frac{\rho V_1}{\rho V_1-170}\\\Rightarrow \frac{m_1}{m_2}=\frac{1.244\times 530}{1.244\times 530-170}=1.347\\\Rightarrow \frac{m_1}{m_2}=1.347\\\Rightarrow \frac{n_1}{n_2}=1.347$

Dividing the first two equations we get

$1=\frac{n_1}{n_2}\frac{T_1}{T_2}\\\Rightarrow 1=1.347\frac{283.15}{T_2}\\\Rightarrow T_2=1.347\times 283.15= 381.52\ K$

∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C

8 0

3 years ago

State Newton's first law of motion .​

State Newton's first law of motion .