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2 years ago

Hai ô tô khởi hành cùng lúc từ A,B cách nhau 20 km chuyển động theo hướng từ A đến B vs vận tốc lần lượt là 60km/h và 40km/h

Physics

1 answer:

sdas [7]2 years ago

6 0

Answer:

áp dụng công thức v = s/t

s là dộ dài qduong

v là vận tốc

t là thời gian

xuyên suốt 2 câu hỏi đều dùng công thức này

Explanation:

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What Causes A Short Circuit?

slavikrds [6]

C. When there is too little resistance and the current increases

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3 years ago

An electron is isolated from an atom and exists in vacuum. A group of scientists collectively state that they can remove part of

lesya692 [45]

D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values

Explanation:

The electric charge of an object is a property of the object that is related to the ability of the object to experience/exert an electric force: if the object is electrically charge, then it is attracted or repelled by other electrically charged object.

The electric charge of an object depends on the amount of charged particles it has on it. In particular, the fundamental particles that carry electric charge are:

Protons: they carry electric charge of +e
Electrons: they carry electric charge of -e

Where "e" is the fundamental charge ( $e=1.6\cdot 10^{-19}C$ ). Therefore, one proton carry a charge of +e and one electron carry a charge of -e.

An electron is a fundamental particle: this means that it cannot be divided into smaller particles. This also means that it is not possible to remove part of the charge of the electron: in fact, it is said that electric charge exists only as discrete values, being a multiple of $e$ . Therefore, the correct statement is

D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values

Learn more about particles:

brainly.com/question/2757829

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5 0

3 years ago

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 44.0 N at the s

BabaBlast [244]

Answer:

Weight at the surface of Jupiter's moon Io is 8.13 N .

Explanation:

Given :

Acceleration due to gravity at the surface of Jupiter's moon is $g_m=1.81\ m/s^2$ .

Weight of watermelon in earth , $W=44\ N$ .

Acceleration due to gravity at the surface of earth is $g=9.81\ m/s^2$ .

We know , weight is given by :

$W=mg\\m=\dfrac{W}{g}\\\\m=4.49\ kg$

Therefore , mass at the surface of Jupiter's moon Io is :

$W_m=mg_m\\\\W_m=4.49\times 1.81\\\\W_m=8.13 \ N$

Hence , this is the required solution .

6 0

3 years ago

Kyle has a mass of 54kg and is moving at 3 m/s what is his kinetic energy

Mars2501 [29]

Answer:

243J

Explanation:

K.E = 1/2 x 54 x 3^2

K.E = 1/2 x 54 x 9

K.E = 1/2 x 486

K.E = 486/2

K.E = 243J

6 0

3 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago