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3 years ago

Một điện tích q = 6.10-6 C đặt tại tâm của một hình lập phương. Tính thông lượng điện trường gửi qua mỗi mặt hình lập phương.

Physics

1 answer:

Lana71 [14]3 years ago

7 0

Answer:

mmmlbdhdjdkekkdnxnfjkckckcklclglglvkglglvkgkvkvkvkvkvkvkvkvkvkvkbkkbkbkkbbkkbkbkbkkbkbkblbkkbkb

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The world speed record on water was set on October

Luda [366]

Answer: Remember speed is distance divided by time, so if he travels 1000 m in 7.045 s, his speed is

(1000 m)/(7.045 s) = 141.9 m/s.

Note there are 1609 metres in a mile, or 1 mi = 1609 m, so m = 1/1609 mi, or

141.9/1609 mi/s = 0.08822 mi/s. Now, note that 1 h = 3600 s, so the speed is

0.08822*3600 mi/h = 317.6 mi/h.

8 0

3 years ago

Three cars with identical engines and tires start from rest, and accelerate at their maximum rate. Car X is the most massive, an

aksik [14]

Answer:C) car X

Explanation:

Given

All the cars have identical Engine thus Force Produced by car X will be equal to Y and Z

and $Force=mass\times acceleration$

Since Car X is most massive so acceleration associated with it will be minimum

acceleration of car X is minimum thus it will travel farthest

6 0

3 years ago

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m$

5 0

3 years ago

What distance will be traveled if you are going 120km/hr for 0.75 hr?

SVEN [57.7K]

Answer:

60 km

Explanation:

For an object (or a person, such as in this case) moving at constant speed, the speed is equal to the ratio between the distance travelled and the time taken:

where

v is the speed

d is the distance

t is the time taken

In this case, we have:

v = 120 km/h is the speed

t = 30 min = 0.5 h is the time taken

Therefore, we can rearrange the equation to find the total distance travelled:

5 0

3 years ago

Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10

11111nata11111 [884]

Answer:

$1.62\times 10^{-8}\ \text{s}$

Explanation:

$\epsilon_0$ = Vacuum permittivity = $8.854\times 10^{-12}\ \text{F/m}$

$A$ = Area = $10\times 2\times 10^{-4}\ \text{m}^2$

$d$ = Distance between plates = 1 mm

$V_c$ = Changed voltage = 60 V

$V$ = Initial voltage = 100 V

$R$ = Resistance = $1000\ \Omega$

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}$

We have the relation

$V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}$

The time taken for the potential difference to reach the required level is $1.62\times 10^{-8}\ \text{s}$ .

5 0

2 years ago