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3241004551 [841]
3 years ago
9

A rocket started from the rest and goes straight up with an acceleration of 4.0 m/s^2 for the first 100 seconds. Then it coasts

forever.
Write a program that asks the user to enter the time since the launch of the rocket in seconds.
Your program should then display:
Entered number in seconds
Speed of the rocket at that time
Altitude of the rocket at that time
Physics
2 answers:
Sonja [21]3 years ago
6 0

Answer:

I think that its is Write a program that ask the user to entaer the time since the launch of the rocket is in seconds sorry if wrong.

Explanation:

Dimas [21]3 years ago
3 0

Answer:

idk

Explanation:

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Please help me with this question​
vovangra [49]

Answer:

1. 12 V

2a. R₁ = 4 Ω

2b. V₁ = 4 V

3a. A = 1.5 A

3b. R₂ = 4 Ω

4. Diagram is not complete

Explanation:

1. Determination of V

Current (I) = 2 A

Resistor (R) = 6 Ω

Voltage (V) =?

V = IR

V = 2 × 6

V = 12 V

2. We'll begin by calculating the equivalent resistance. This can be obtained as follow:

Voltage (V) = 12 V

Current (I) = 1 A

Equivalent resistance (R) =?

V = IR

12 = 1 × R

R = 12 Ω

a. Determination of R₁

Equivalent resistance (R) = 12 Ω

Resistor 2 (R₂) = 8 Ω

Resistor 1 (R₁) =?

R = R₁ + R₂ (series arrangement)

12 = R₁ + 8

Collect like terms

12 – 8 =

4 = R₁

R₁ = 4 Ω

b. Determination of V₁

Current (I) = 1 A

Resistor 1 (R₁) = 4 Ω

Voltage 1 (V₁) =?

V₁ = IR₁

V₁ = 1 × 4

V₁ = 4 V

3a. Determination of the current.

Since the connections are in series arrangement, the same current will flow through each resistor. Thus, the ammeter reading can be obtained as follow:

Resistor 1 (R₁) = 4 Ω

Voltage 1 (V₁) = 6 V

Current (I) =?

V₁ = IR₁

6 = 4 × I

Divide both side by 4

I = 6 / 4

I = 1.5 A

Thus, the ammeter (A) reading is 1.5 A

b. Determination of R₂

We'll begin by calculating the voltage cross R₂. This can be obtained as follow:

Total voltage (V) = 12 V

Voltage 1 (V₁) = 6 V

Voltage 2 (V₂) =?

V = V₁ + V₂ (series arrangement)

12 = 6 + V₂

Collect like terms

12 – 6 = V₂

6 = V₂

V₂ = 6 V

Finally, we shall determine R₂. This can be obtained as follow:

Voltage 2 (V₂) = 6 V

Current (I) = 1.5 A

Resistor 2 (R₂) =?

V₂ = IR₂

6 = 1.5 × R₂

Divide both side by 1.5

R₂ = 6 / 1.5

R₂ = 4 Ω

4. The diagram is not complete

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2 years ago
A 2.00 kg block is placed against a spring on a frictionless 36° incline. the spring, whose spring constant is 19.8n/cm, is com
alekssr [168]
A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs. 
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5 0
3 years ago
What is the energy released in this B- nuclear reaction 2K-> 2Ca0,e? (The atomic mass of 42 K is 41.962403 u and that of 42Ca
Katyanochek1 [597]

<u>Answer:</u> The energy released in the given nuclear reaction is 3.526 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

_{19}^{42}\textrm{K}\rightarrow _{20}^{42}\textrm{Ca}+_{-1}^{0}\textrm{e}

We are given:

Mass of _{19}^{42}\textrm{K} = 41.962403 u

Mass of _{20}^{42}\textrm{Ca} = 41.958618 u

To calculate the mass defect, we use the equation:

\Delta m=\text{Mass of reactants}-\text{Mass of products}

Putting values in above equation, we get:

\Delta m=(41.962403-41.958618)=0.003785u

To calculate the energy released, we use the equation:

E=\Delta mc^2\\E=(0.003785u)\times c^2

E=(0.003785u)\times (931.5MeV) (Conversion factor: 1u=931.5MeV/c^2 )

E=3.526MeV

Hence, the energy released in the given nuclear reaction is 3.526 MeV.

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3 years ago
A car of mass 1200 kg, moving with a speed of 72 mph on a highway,
Keith_Richards [23]

momentum(p) = mass(m) x velocity(v)

KE = kinetic energy = 1/2 mv²

a.

the ratio of the momentum of the SUV to that of the car

\tt \dfrac{mv}{1.5m\dfrac{2}{3}v }=1:1

b.  the ratio of the KE of the SUV to that of the car

\tt \dfrac{1/2\times mv^2}{1/2\times 1.5m\times (\dfrac{2}{3}v )^2}=\dfrac{mv^2}{1.5m\dfrac{4}{6}v^2 }=1:1

8 0
1 year ago
Jump to Question: Which of these is not required for current to flow through a conductor? A. Source of electrical current B. Som
amid [387]
I believe it is D. something with air molecules
7 0
3 years ago
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