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2 years ago

How do you build a sticky piston door

Physics

2 answers:

Pani-rosa [81]2 years ago

7 0

Answer:

it can be built in mine craft

Explanation:

IRISSAK [1]2 years ago

7 0

Answer:

put 2 pistons on each side and then put the block you want the door to be and power it with redstone

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What fundimental force is responsible for the repulsion between two positively charged particles?

Aliun [14]

Coulomb interaction is responsible

4 0

3 years ago

What decibel reading corresponds to a pressure amplitude of 0.2 W/m^2?

horsena [70]

I = pressure amplitude given = 0.2 W/m²

dB = decibel reading

decibel reading from the pressure amplitude is given as

dB = 10 log₁₀ (I/10⁻¹²)

inserting the values in the above equation

dB = 10 log₁₀ (0.2/10⁻¹²)

dB = 10 log₁₀ (2 x 10⁻¹/10⁻¹²)

dB = 10 log₁₀ (2 x 10⁻¹.10¹²)

dB = 10 log₁₀ (2 x 10¹²⁻¹)

dB = 10 log₁₀ (2 x 10¹¹)

dB = 113.01 db

hence the decibel reading comes out to be 113.01 db

4 0

3 years ago

Cart A and cart B are traveling in the same direction on a straight track. Cart A is moving at 9.25 m/s and cart B is moving at

Liula [17]

Answer:

The speed of cart B is 11.21 m/s.

Explanation:

Given that,

Speed of cart A = 9.25 m/s

Speed of cart B = 7.15 m/s

Mass of cart A = 72.0 kg

Mass of cart B = 55.0 kg

Speed of card A after collision = 6.15 m/s

We need to calculate the speed of cart B

Using conservation of momentum

$m_{A}v_{A}+m_{B}v_{B}=m_{A}v_{A}+m_{B}v_{B}$

Put the value into the formula

$72.0\times9.25+55.0\times7.15=72.0\times6.15+55.0\times v_{B}$

$v_{B}=\dfrac{72.0\times9.25+55.0\times7.15-72.0\times6.15}{55.0}$

$v_{B}=11.21\ m/s$

Hence, The speed of cart B is 11.21 m/s.

7 0

2 years ago

• A cable is suspended over a pulley. A 5.0 kg mass is attached to

Galina-37 [17]

Answer:

Tension= $\frac{20g}{3}$ (g=acceleration of gravity)

Explanation:

Given that,

A 5Kg and 10Kg are attached by a cable suspended over a pulley.

As 10Kg > 5Kg , the 10 kg mass accelerates down and the 5kg mass accelerates up, let it be a. Let the tension in the cable be T.

So, the equations of motion are

$10g-T = 10a$

$T-5g=5a$

Now adding them we get,

$5g=15a$

$a=\frac{g}{3}$

Substituting them back in the equation we get,

$10g-10(\frac{g}{3} ) = T$

$T= \frac{20g}{3}$

3 0

3 years ago

Force F → = (−8.0 N)iˆ + (6.0 N)jˆ acts on a particle with position vector r → = (3.0 m)iˆ + (4.0 m)jˆ. What are (a) the torque

disa [49]

Answer with Explanation:

We are given that

$F=-8\hat{i}+6\hat{j}$

$r=3\hat{i}+4\hat{j}$

a.We have to find the torque on the particle about the origin.

We know that

Torque= $\tau=r\times F=\begin{vmatrix}i&j&k\\3&4&0\\-8&6&0\end{vmatrix}$

By using the formula

$\tau=50\hat{k}$

b. $\mid \tau\mid =\mid F\mid \mid r\mid sin\theta$

$\mid F\mid=\sqrt{(-8)^2+(6)^2}=10$

$\mid r\mid=\sqrt{3^2+4^2}=5$

$\mid \tau\mid=\sqrt{(-50)^2}=50$

Substitute the values then we get

$50=10\times 5 sin\theta$

$sin\theta=\frac{50}{50}=1$

$sin\theta=sin90^{\circ}$

Because $sin90^{\circ}=1$

$\theta=90^{\circ}$

3 0

3 years ago

How do you build a sticky piston door​

How do you build a sticky piston door