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melisa1 [442]
3 years ago
10

6. One minute after takeoff, a rocket carrying the space shuttle into outer space reaches a speed of 447 m/s.

Physics
1 answer:
Sidana [21]3 years ago
7 0

Answer:

acceleration of the rocket is given as

a = 7.45 m/s^2

Explanation:

As we know that rocket starts from rest and then reach to final speed of 447 m/s after t = 1 min

so we have

v_i = 0

v_f = 447 m/s

t = 1 min = 60 s

so we have

a = \frac{v_f - v_i}{t}

a = \frac{447 - 0}{60}

a = 7.45 m/s^2

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san4es73 [151]

Momentum = mass x velocity

Thus Option A is the correct answer

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= 134.1 Kg m/s

Momentum ( bullet)  = 0.02 kg x (0.447 x 800) m/s

= 7.152 Kg m/s

Momentum ( truck) = 0, as v = 0

tightrope has both low mass  and low speed, thus its momentum will be low


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How much force is required to accelerate a 2kg mass at 3 m/s2? answer key?
mr Goodwill [35]
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4 0
3 years ago
How do two sublevels of the same principal energy level differ from each other
Colt1911 [192]

Each energy sublevel corresponds to an orbital of a different shape.

Explanation:

Two sublevels of the same principal energy level differs from each other if the sublevels corrresponds to an orbital of a different shape.

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Take for example, Carbon:

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7 0
3 years ago
One object (m1 = 0.220 kg) is moving to the right with a speed of 2.10 m/s when it is struck from behind by another object (m2 =
blagie [28]

Answer:

vf₁  = 6.86 m/s , to the right

vf₂ =  2.96 m/s, to the right

Explanation:

Theory of collisions  

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:  

p=m*v  

where  

p:Linear momentum  

m: mass  

v:velocity  

There are 3 cases of collisions : elastic, inelastic and plastic.  

For the three cases the total linear momentum quantity is conserved:  

P₀ = Pf Formula (1)  

P₀ :Initial linear momentum quantity  

Pf : Final linear momentum quantity  

Data

m₁= 0.220 kg : mass of  object₁

m₂= 0.345 kg : mass of  object₂

v₀₁ =  2.1 m/s ₁ , to the right : initial velocity of m₁

v₀₂=   6 m/s, to the right  i :initial velocity of m₂

Problem development

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂  

We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:

(0.22)*(2.1) + (0.345)*(6) = (0.22)*vf₁ +(0.345)*vf₂

2.532 = (0.22)*vf₁ +(0.345)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

e= \frac{v_{f2}-v_{f1} }{v_{o1} -v_{o2} }

1*(v₀₁ - v₀₂ )  = (vf₂ -vf₁)

(2.1 - 6 )  = (vf₂ -vf₁)

-3.9 =  (vf₂ -vf₁)

vf₂ = vf₁ - 3.9

vf₂ = vf₁ - 3.9 Equation (2)

We replace Equation (2) in the Equation (1)

2.532 = (0.22)*vf₁ +(0.345)*( vf₁ - 3.9)

2.532 = (0.22)*vf₁ +(0.345)* (vf₁) -(0.345)( 3.9)

2.532 + 1.3455 = (0.565)*vf₁

3.8775 = (0.565)*vf₁

vf₁  = (3.8775) / (0.565)

vf₁  = 6.86 m/s, to the right

We replace vf₁  = 6.86 m/s in the Equation (2)

vf₂ =  6.86 - 3.9

vf₂ =  2.96 m/s, to the right

8 0
3 years ago
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