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2 years ago

When Bob needs to send Alice a message with a digital signature, whose private key is used to encrypt the hash?

Physics

1 answer:

nadya68 [22]2 years ago

3 0

Answer: Bob's private key

Explanation:

Since Bob is sending Alice the message, it simply implies he must have encrypted it with his key that Alice can be able to use to decrypt the message.

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in a softball game, a batter hits the ball at the velocity of 27m/s and angle of 40 shown below. What is the maximum range of th

Nata [24]

Answer:

R = 73.25 m

Explanation:

We have,

Initial speed of the ball is 27 m/s

It is projected at an angle of 40 degrees

The maximum range of the ball is given by :

$R=\dfrac{u^2\sin2\theta}{g}$

Plugging all the values we get :

$R=\dfrac{(27)^2\sin2(40)}{9.8}\\R=73.25\ m$

So, the maximum range of the ball is 73.25 m

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2 years ago

What is the common phrase for natural selection

Rainbow [258]

Natrual selection is where animalz better adapt to their enviroment

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2 years ago

A balloon having an initial temperature of 17.8Â°C is heated so that the volume doubles while the pressure is kept fixed. What i

Rama09 [41]

Answer:

$T = 308.6 ^0 C$

Explanation:

Here by ideal gas equation we can say

$PV = nRT$

now we know that pressure is kept constant here

so we will have

$V = \frac{nR}{P} T$

since we know that number of moles and pressure is constant here

so we have

$\frac{V_2}{V_1} = \frac{T_2}{T_1}$

now we know that initial temperature is 17.8 degree C

and finally volume is doubled

So we have

$\frac{2V}{V} = \frac{T_2}{(273 + 17.8)}$

so final temperature will be

$T_2 = 581.6 k$

$T_2 = 308.6 ^o C$

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2 years ago

What are the three different types of friction

iogann1982 [59]

Answer:

sliding , static , rolling

Explanation:

HOPE IT WILL HELP YOU

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2 years ago

A rocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s

topjm [15]

U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s

Let h₁ = the height at time t, for t ≤ 3.98 s
Let h₂ = the height at time t > 3.98 s

Motion for t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s

Motion for t > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m

The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m

Answer: 931.4 m (nearest tenth)

6 0

3 years ago

Read 2 more answers