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Schach [20]
3 years ago
6

A 4.0 kg model rocket is launched, shooting 50.0 g of burned fuel from its exhaust at

Physics
1 answer:
Anettt [7]3 years ago
3 0

The velocity of the rocket is 7.8 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the rocket+fuel system must be conserved.

Before the launch, the total momentum of the system is zero, since the rocket and the fuel are at rest:

p=0

After the launch, the total momentum is

p=MV+mv

where

M = 4.0 kg is the mass of the rocket

V is the velocity of the rocket

m = 50.0 g = 0.050 kg is the mass of the fuel ejected

v = -625 m/s is the velocity of the fuel (taking "backward" as negative direction)

Since the total momentum is conserved, we have

0=MV+mv

So we can solve the equation to find V, the velocity of the rocket:

V=-\frac{mv}{M}=-\frac{(0.050)(625)}{4.0}=+7.8 m/s

And the positive sign means the rocket moves forward.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc
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Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

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Answer:

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Explanation:

The internal axial load at point D can be calculated as the change in the subjected loads. if the magnitude of the horizontal direction = zero

EF_x = 0; Then:

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