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fgiga [73]
3 years ago
10

Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims

that it will maintain a constant thrust of 12.3 N 12.3 N until the engine is used up. Robert launches the rocket on a windless day, so that it travels straight up, and uses his laser range‑finder to meaure that the height of the rocket when the engine cuts off is 10.2 m 10.2 m . He also measures the rocket's peak height, which is 14.7 m 14.7 m . If the rocket has a mass of 0.663 kg 0.663 kg , how much work is done by the drag force on the rocket during its ascent?
Physics
1 answer:
777dan777 [17]3 years ago
5 0

Answer:

The work done by the drag force is given by 29.96 J

Explanation:

Given :

Thrust force F = 12.3 N

Displacement d = 10.2 m

Mass of rocket m = 0.663 Kg

From work energy theorem,

  W = \Delta K

 W_{t} - Wd - W_{g} = KE

Where W_{t} = thrust work W_{g} = gravitational work

KE = 12.3 \times 10.2 -Wd - 0.663 \times 9.8 \times 10.2

KE = 59.2 -Wd

After cutoff kinetic energy is converted into potential energy,

KE = Wd' + mg\Delta h

Put value of KE

59.2 -Wd = Wd' + 0.663 \times 9.8 \times 4.5

Work done by drag force is given by,

Wd'+Wd =  59.2 -29.23

                 = 29.96 J

Therefore, the work done by the drag force is given by 29.96 J

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Mass m moves to the right with speed =v along a frictionless horizontal surface and crashes into an equal mass m initially at re
Amiraneli [1.4K]

After the collision the magnitude of the momentum of the system is Mv

Given:

mass of 1st object = M

speed of 1st object = v

mass of 2nd object = M

speed of 2nd object = 0

To Find:

magnitude of the momentum after collision

Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.

Applying conservation of linear momentum

Mv + M(0) = 2MV

Mv = 2MV

V = v/2

So, after collision momentum is

p = 2MV = 2xMxv/2 = Mv

So, after collision momentum is Mv

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4 0
1 year ago
A car changes speed from 25 m/s to 10 m/s in 240 seconds. Describe its acceleration.​
Dmitry_Shevchenko [17]

Explanation:

Acceleration is change in velocity over change in time:

a = Δv / Δt

a = (10 m/s - 25 m/s) / (240 s - 0 s)

a = -0.0625 m/s²

So the car decelerates at 0.0625 m/s².

7 0
3 years ago
Calculate the change in entropy as 0.3071 kg of ice at 273.15 K melts. (The latent heat of fusion of water is 333000 J / kg)
Ostrovityanka [42]

Answer:

374.39 J/K

Explanation:

Entropy: This can be defined as the degree of disorder or randomness of a substance.

The S.I unit of entropy is J/K

ΔS = ΔH/T ..................................... Equation 1

Where ΔS = entropy change, ΔH = Heat change, T = temperature.

ΔH = cm................................... Equation 2

Where,

c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.

Substitute into equation 2

ΔH = 333000×0.3071

ΔH = 102264.3 J.

Also, T = 273.15 K

Substitute into equation 1

ΔS = 102264.3/273.15

ΔS = 374.39 J/K

Thus, The change in entropy = 374.39 J/K

3 0
2 years ago
Determine (a) the starting height, (b) the time to hit the ground, and (c) the velocity when it hits the ground for an object sh
Salsk061 [2.6K]

Answer:

a

Explanation:

8 0
2 years ago
A 20-lb force acts to the west while an 80-lb force acts 45° east of north. The magnitu 80 lb 70 lb 067 lb 100 lb
natulia [17]

Answer:

option C

Explanation:

given,

force act on west  = 20 lb

force act at 45° east of north = 80 lb

magnitude of force = ?

∑ F y  =  80 cos 45⁰

    F y =  56.57 lb

magnitude of forces in x- direction

∑ F x = -20 + 80 sin 45⁰

        = 36.57 lb

net force

F = \sqrt{F_x^2+F_y^2}

F = \sqrt{56.57^2+36.57^2}

F = 67.36 lb≅ 67 lb

hence, the correct answer is option C

4 0
3 years ago
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