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VARVARA [1.3K]
3 years ago
11

Please help!!! This Is for science

Physics
1 answer:
faust18 [17]3 years ago
7 0

Answer:

it is a

Explanation:

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What is the power involved in lifting a 1.0-kg object 1.0 m in 1.0 s??
MAVERICK [17]
Mass  = 1kg 
Distance  = 1m 
Time = 1s
Force= Mass x Acceleration due to graviy
          =  1 x 9.8 = 9.8
Velocity = Distance  / time
              = 1 / 1 =1m/s
Power = Force x velocity
           = 9.8 x 1 = 9.8 W
5 0
4 years ago
A jet aircraft is traveling at 262 m/s in horizontal flight. The engine takes in air at a rate of 85.9 kg/s and burns fuel at a
Fittoniya [83]

Answer:

F_T=60132.52N

P=15814852.76W

Explanation:

From the question we are told that

Velocity of aircraft  V=263m/s

Engine air intake rate \triangle M_a=85.9kg/s

Fuel burn rate  \triangle M_f =3.92kg/s

Velocity of exhaust gas V_e =921m/s

Generally the Mass change rate of Rocket is mathematically given by

 \triangle M = \triangle M_a+\triangle M_f

 \triangle M= 85.9+3.92

 \triangle M=89.82kg/s

Generally the Trust of the rocket is given mathematically by

 F_T=(\triangle M *V_e)-(dM_a/dt)*(V)

 F_T=(89.82 *921)-(85.9)*(263)

 F_T=60132.52N

Generally the Rocket's delivered power is mathematically given by

Delivered power P

 P=V*F_T

 P=263*60132.52N

 P=15814852.76W.

8 0
3 years ago
What kind of energy does a rubber band have when it is stretched?
VashaNatasha [74]
The answer is obviously a because a rubber band uses elastic energy
3 0
4 years ago
An airplane is flying west at 200km/h speeds up to 300km/h over 2 hours. what is the acceleration
Andreas93 [3]

Answer:

600 km/h

Explanation:

3 0
3 years ago
A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = \frac{I}{nqA}

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

v = \frac{I}{nqA}

v = \frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0
4 years ago
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