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elena55 [62]
2 years ago
8

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

You might be interested in
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
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