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mezya [45]
3 years ago
6

The pressure less than atmospheric pressure is known as:

Engineering
1 answer:
Ber [7]3 years ago
3 0

Answer:

answer is option (d) all of the above

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section 10.1 describes a man in the middle attack on the diffie-hellman key exhange protocal in which the adversary generates tw
nika2105 [10]

Answer:

1. James will attack by generating a random private key XD and a corresponding public key YD.

2. Jane transmit YA to another person called Alex.

3. James intercept YA and transmit YD to jane.

4. Jane receive YD and calculate K1

At this point james and jane thinks they share a secret key but instead james has a secret key k1 to Jane and k2 to alex.

5. Alex transmit another key XA to alex for example.

6. James intercept and calculate k2 and vice versa.

6 0
3 years ago
You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19
Pepsi [2]

Answer:

<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

V = A*t

<u>Where</u>:

A: is the surface area = 160

t: is the thickness = 0.002

<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>

V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}    

Now, using the density we can find the mass:

m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g

Finally, with the Avogadros number (N_{A}) and with the atomic mass (A) we can find the number of atoms (N):

N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0
3 years ago
Pumped-storage hydroelectricity is a type of hydroelectric energy storage used by electric power systems for load balancing. The
NikAS [45]

Answer:

A) energy loss E = pgQtH

Where p = density in kg/m3

g = gravity acceleration in m/s2

Q = flow rate in m3/s

t = time taken for flow in sec

H = height of flow in m

B) power required to run pump;

P = pgQH

Explanation:

Detailed explanation and calculation is shown in the image below

5 0
3 years ago
What combustion after effects do actuators work to control
Rufina [12.5K]
Answer: I would help you but I don’t know the answer, so sorry
4 0
3 years ago
Read 2 more answers
A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c
Paraphin [41]

Answer:

10.8\ \text{lb/ft^2}

101.96\ \text{lb/ft}^2

Explanation:

v_1 = Velocity of car = 65 mph = 65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}

\rho = Density of air = 0.00237\ \text{slug/ft}^3

v_2=0

P_1=0

h_1=h_2

From Bernoulli's law we have

P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}

The maximum pressure on the girl's hand is 10.8\ \text{lb/ft^2}

Now v_1 = 200 mph = 200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}

P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2

The maximum pressure on the girl's hand is 101.96\ \text{lb/ft}^2

5 0
3 years ago
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