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Aleksandr [31]
3 years ago
8

Define oxidation number

Physics
1 answer:
Delicious77 [7]3 years ago
8 0

<span>a number assigned to an element in a chemical combo that represents the number of electrons lost or gained by atom of the element in the compound.</span>

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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 m /
Sidana [21]

Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 7.00 s

We need to calculate the distance

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value into the formula

s=0+\dfrac{1}{2}\times29.4\times7^2

s=720.3\ m

We need to calculate the velocity

Using formula of velocity

v=a\times t

Put the value into the formula

v=29.4\times7

v=205.8\ m/s

We need to calculate the height

Using formula of height

H=\dfrac{v^2}{2g}

Put the value into the formula

H=\dfrac{(205.8)^2}{2\times9.8}

H=2160.9\ m

We need to calculate the maximum height

Using formula for maximum height

H'=H+s

Put the value into the formula

H'=2160.9+720.3

H'=2881.2\ m

Hence, The maximum height is 2881.2 m.

4 0
3 years ago
Mechanical energy is the sum of kinetic and potential energy in an object. It is energy in an object due to its motion, position
german
Mechanical energy is made when something is moved. The energy that is moving is kinetic. And potential energy is stored energy. Mechanical energy can be used to store energy and to cause moving energy. For instance: a slingshot. Pulling back the band creates potential energy and releasing it creates kinetic energy.
6 0
3 years ago
Read 2 more answers
You have a grindstone (a disk) that is 95.2 kg, has a 0.399 m radius, and is turning at 93 rpm, and you press a steel axe agains
olya-2409 [2.1K]

Answer:

angular acceleration is -0.2063  rad/s²

Explanation:

Given data

mass m = 95.2 kg

radius r = 0.399 m

turning ω = 93 rpm

radial force N  = 19.6 N

kinetic coefficient of friction  μ = 0.2

to find out

angular acceleration

solution

we know frictional force that is = radial force × kinetic coefficient of friction

frictional force = 19.6 × 0.2

frictional force = 3.92 N

and

we know moment of inertia  that is

γ =  I ×α = frictional force × r

so

γ  = 1/2 mr²α

α  = -2f /mr

α  = -2(3.92) /95.2 (0.399)

α  = - 7.84 / 37.9848 = -0.2063

so angular acceleration is -0.2063  rad/s²

3 0
3 years ago
During the middle of a family picnic, Barry Allen received a message that his friends Bruce and Hal
weeeeeb [17]

The kinematics of the uniform motion and the addition of vectors allow finding the results are:

  • The  Barry's initial trajectory is 94.30 10³ m with n angles of θ = 138.8º
  • The return trajectory and speed are v = 785.9 m / s, with an angle of 41.2º to the South of the East

Vectors are quantities that have modulus and direction, so they must be added using vector algebra.

A simple method to perform this addition in the algebraic method which has several parts:

  • Vectors are decomposed into a coordinate system
  • The components are added
  • The resulting vector is constructed

 Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case

              v = \frac{\Delta d}{t}

              Δd = v t

Where  v is the average velocity, Δd the displacement and t the time

We look for the first distance traveled at speed v₁ = 600 m / s for a time

          t₁ = 2 min = 120 s

          Δd₁ = v₁ t₁

          Δd₁ = 600 120

          Δd₁ = 72 10³ m

Now we look for the second distance traveled for the velocity v₂ = 400 m/s    

  time t₂ = 1 min = 60 s

          Δd₂ = v₂ t₂

          Δd₂ = 400 60

          Δd₂ = 24 103 m

   

In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,

We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)

Let's use trigonometry for each distance

Route 1

          cos (180 -35) = \frac{x_1}{\Delta d_1}

          sin 145 = \frac{y_1}{\Delta d1}

          x₁ = Δd₁ cos 125

          y₁ = Δd₁ sin 125

          x₁ = 72 103 are 145 = -58.98 103 m

          y₁ = 72 103 sin 155 = 41.30 10³ m

Route 2

          cos (90+ 30) = \frac{x_2}{\Delta d_2}

          sin (120) = \frac{y_2}{\Delta d_2}

          x₂ = Δd₂ cos 120

          y₂ = Δd₂ sin 120

          x₂ = 24 103 cos 120 = -12 10³ m

           y₂ = 24 103 sin 120 = 20,78 10³ m

             

The component of the resultant vector are

              Rₓ = x₁ + x₂

              R_y = y₁ + y₂

              Rx = - (58.98 + 12) 10³ = -70.98 10³ m

              Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m

We construct the resulting vector

Let's use the Pythagoras' Theorem for the module

             R = \sqrt{R_x^2 +R_y^2}

             R = \sqrt{70.98^2 + 62.08^2}   10³

             R = 94.30 10³ m

We use trigonometry for the angle

             tan θ ’= \frac{R_y}{R_x}

             θ '= tan⁻¹ \frac{R_y}{R_x}

             θ '= tan⁻¹ \frac{62.08}{70.98}

             θ ’= 41.2º

Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction

            θ = 180 - θ'

            θ = 180 -41.2

            θ = 138.8º

Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.

The average speed of the trip should be

             v = \frac{\Delta R}{t_3}  

             v = \frac{94.30}{120}  \ 10^3

              v = 785.9 m / s

in the opposite direction, that is, the angle must be

               41.2º to the South of the East

In conclusion, using the kinematics of the uniform motion and the addition of vectors, results are:

  • To find the initial Barry trajectory is 94.30 10³ m with n angles of  138.8º
  • The return trajectory and speed is v = 785.9 m / s, with an angle of 41.2º to the South of the East

Learn more here:  brainly.com/question/15074838

4 0
2 years ago
A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s2. Pa
ValentinkaMS [17]

Answer:

45 s .

Explanation:

The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .

displacement s = ?

acceleration a = 1 m /s²

Final speed v = 5 m/s

initial speed u = 0

v² = u² + 2as

5² = 0 + 2 x 1 x s

s = 12.5  m

B)  Let time of acceleration or deceleration be t

v = u + a t

5 = 0 + 1 t

t = 5 s

Similarly displacement during deceleration = 12.5 m

Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) =  175 m .

velocity of uniform motion = 5 m /s

time during which there was uniform velocity = 175 / 5 = 35 s

Total time = 5 + 35 + 5 = 45 s .

4 0
2 years ago
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