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2 years ago

Which subatomic particle has the smallest mass?

Physics

1 answer:

spin [16.1K]2 years ago

3 0

Answer:

quark

Explanation:

neutrón= 1.675 x 10-27 kg

proton=1,673 x 10-27 kg

electron=9,11 x 10-31 kg

quark=7,13 - 14,26 · 10−30 kg

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An object begins with a speed of 20 meters per second and slows down to a speed of 10 meters per second over a time of 5 seconds

Andreyy89

Answer:

V = 6 m/s

Explanation:

Given that,

Initial speed of an object is 20 m/s

Final speed of an object is 10 m/s

Time, t = 5 s

We need to find the average speed of the object during these 5 seconds. Let it is equal to V. Here, time is same. The average speed is given by :

$V=\dfrac{x+y}{2}\\\\\text{Putting values}\\\\V=\dfrac{20+10}{5}\\\\V=6\ m/s$

So, the average speed of the object is 6 m/s.

3 0

3 years ago

Water ﬂows through a horizontal nozzle in steady ﬂow at the rate of 10m3/s. The inlet and outlet diameters are d1 = 0.5m and d2

Dvinal [7]

Answer:

P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,

Explanation:

This problem of fluid mechanics let's start with the continuity equation to find the speed of water output

Q = A v

v = Q / A

The area of a circle is

A = π r² = π d² / 4

Let's look at the speeds at each point

v₁ = Q / A₁ = Q 4 /π d₁²

v₁ = 10 4 /π 0.5²

v₁ = 50.93 m / s

v₂ = Q / A₂

v₂ = 10 4 /π 0.25²

v₂ = 203.72 m / s

Now we can use Bernoulli's equation in the colon

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear

P₁ = P2 + ½ rho (v₂² - v₁²)

P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)

P₁ = 1.013 10⁵ + 2.205 10⁷

P₁ = 2.215 10⁷ Pa

la definicion de presion es

P₁ = F₁/A₁

F₁ = P₁ A₁

F₁ = 2.215 10⁷ pi d₁²/4

F₁ = 2.215 10⁷ pi 0.5²/4

F₁ = 4.3 106 N

6 0

3 years ago

Can someone please help me with this question and I will give you brainlest

grigory [225]

Answer:

PEENNISS

Explanation:

7 0

3 years ago

Read 2 more answers

the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr

Dmitry [639]

Answer:

The approximate number of decays this represent is $N= 23*10^{10}$

Explanation:

From the question we are told that

The amount of Radiation received by an average american is $I_a = 2.28 \ mSv$

The source of the radiation is $S = 5.49 MeV \ alpha \ particle$

Generally

$1 \ J/kg = 1000 mSv$

Therefore $2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg$

Also $1eV = 1.602 *10^{-19}J$

Therefore $2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg} * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg$

An Average american weighs 88.7 kg

The total energy received is mathematically evaluated as

$1 kg ------> 1.423*10^{16}ev \\88.7kg --------> x$

Cross-multiplying and making x the subject

$x = 88.7 * 1.423*10^{16} eV$

$x = 126.2*10^{16}eV$

Therefore the total energy deposited is $x = 126.2*10^{16}eV$

The approximate number of decays this represent is mathematically evaluated as

N = $\frac{x}{S}$

Where n is the approximate number of decay

Substituting values

$N = \frac{126 .2*10^{16}}{5.49*10^6}$

$N= 23*10^{10}$

7 0

4 years ago

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

svetoff [14.1K]

Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Explanation:

Given that $L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V$ , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

$E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}$

#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Hence the charge at any time, t is $q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

6 0

4 years ago