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astraxan [27]
2 years ago
14

A Rocket is launched and reaches a height of 72m before falling back to Earth a) What was it's take off velocity? b) What was it

's final velocity when it landed? c) What was it's velocity at the highest point?​
Physics
1 answer:
DerKrebs [107]2 years ago
4 0

Answer:

Answers are in the explanation

Explanation:

a)

vf² = vi² + 2gh

0 = vi² + 2(-9.8)(72)

vi = 37.57 m/s

b)

vf = vi + gt

0 = 37.57 - (9.8)t

t = 3.83 s

3.83 x 2 = 7.67 s

vf = vi + gt

vf = 37.57 - (9.8)(7.67)

vf = -37.60 m/s

c)

Its vertical velocity at the highest point is 0 m/s

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What is one watt ? Write the relation of watt with kilowatt , megawatt , and horsepower .​
alina1380 [7]

Answer:

See the explanation below

Explanation:

The watt (the power) is equal to the relationship between the work and the time in which that work is performed.

P = W/t

where:

W = work [J] (units of Joules)

t = time [s].

Now 1000 [W] are equal to 1 [kW]

And 1000000 [W] are equal to 1 [MW]

The horsepower is the unit of power in the imperial system of units.

And 745.7 [W] are equal to 1 [Hp]

3 0
2 years ago
A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas
Nookie1986 [14]

The maximum amount of work performed is

W_{max}=\frac{T_H-T_C}{T_C}Q_C

Explanation:

The efficiency of a real heat engine is given by the equation:

\eta = 1-\frac{T_C}{T_H} (1)

where

T_C is the temperature of the cold reservoir

T_H is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

\eta = \frac{W_{max}}{Q_H}

where

W_{max} is the maximum work done

Q_H is the heat absorbed from the hot reservoir

Q_H can be written as

Q_H=W_{max}+Q_C

where

Q_Cis the heat released to the cold reservoir

So the previous equation can be also written as

\eta=\frac{W_{max}}{W_{max}+Q_C} (2)

By combining eq.(1) and (2) we get

1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}

And re-arranging the equation and solving for W_{max}, we find

W_{max}=\frac{T_H-T_C}{T_C}Q_C

Learn more about work and heat:

brainly.com/question/4759369

brainly.com/question/3063912

brainly.com/question/3564634

#LearnwithBrainly

8 0
3 years ago
Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of ea
dimaraw [331]

Answer:

stone A is diamond.

Explanation:

given,

Volume of the two stone =  0.15 cm³

Mass of stone A = 0.52 g

Mass of stone B = 0.42 g

Density of the diamond =  3.5 g/cm³

So, to find which stone is gold we have to calculate the density of both the stone.

We know,

density\density = \dfrac{mass}{volume}

density of stone A

\rho_A = \dfrac{0.52}{0.15}

\rho_A = 3.467\ g/cm^3

density of stone B.

\rho_B = \dfrac{0.42}{0.15}

\rho_B = 2.8\ g/cm^3

Hence, the density of the stone A is the equal to Diamond then stone A is diamond.

6 0
2 years ago
A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If
ella [17]

Answer:

442.5 rad

Explanation:

w₀ = initial angular velocity of the disk = 7.0 rad/s

α = Constant angular acceleration = 3.0 rad/s²

t = time period of rotation of the disk = 15 s

θ = angular displacement of the point on the rim

Angular displacement of the point on the rim is given as

θ = w₀ t + (0.5) α t²

inserting the values

θ = (7.0) (15) + (0.5) (3.0) (15)²

θ = 442.5 rad

4 0
2 years ago
Why is the answer C?
4vir4ik [10]

Explanation:

We want to find the statement that is proven by the fact that the balls reach the same height.

A isn't supported by the evidence.  Balls can reach the same height without having the same initial speed.

B isn't supported by the evidence.  Balls can reach the same height without having the same launch angle.

C is supported.  Projectiles spend the same amount of time going up as they do coming down, so if two projectiles reach the same height, then they must spend the same amount of time in the air.

D isn't supported by the evidence.  Balls thrown at the same speed and complementary angles have the same range but different heights.

E isn't supported by the evidence.  The mass of the ball doesn't affect the height.

7 0
3 years ago
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