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BabaBlast [244]
3 years ago
10

A 62 kg astronaut is space-walking outside the space capsule and is stationary when the tether line breaks. As a means of return

ing to the capsule, he throws his 3.3 kg wrench at a speed of 20.5 m/s away from the capsule. At what speed does the astronaut move toward the capsule
Physics
2 answers:
DanielleElmas [232]3 years ago
8 0

To solve this problem we must apply the concepts related to the conservation of momentum. We have the two masses of the bodies and the velocity of one of those bodies. Therefore, the initial momentum is equivalent to the final momentum. Mathematically this can be expressed as

P_i = P_f

m_1_v_1 = m_2 v_2

Replacing with our values,

3.3kg(20.5m/s) = (62kg)v_2

Solving for v_2,

v_2= 1.09m/s

Therefore the speed that the astronaut move toward the capsule is 1.09m/s

Leokris [45]3 years ago
3 0

Answer:

Explanation:

Given that,

Mass of astronaut

M = 62kg

Mass of wrench

m = 3.3kg

Speed at which wrench move away from the capsule

v = 20.5m/s

Speed of astronaut?

Let speed of astronaut be V

Since there is no external force, we would apply conservation of momentum

Momentum of the astronaut is equal to the momentum of the wrench

Momentum is given as

P = Mass × Velocity

P(astronaut) = P(wrench)

M×V = m×v

V = m×v/M

V = 3.3×20.5/62

V = 1.09 m/s

The astronaut speed is 1.09m/s

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Scilla [17]

Answer:

Option C is correct.

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Explanation:

For combination of materials, the properties (especially physical properties) of the resulting composite is a sum of the fractional contribution of each material thay makes up the composite.

In this composite,

The fibres = 20 vol%

Aluminium = 80 vol%

Modulus of elasticity of the composite

= [0.2 × E(fibres)] + [0.8 × E(Al)]

Modulus of elasticity of the fibers = E(fibres) = (55 × 10⁶) psi. =

Modulus of elasticity of aluminum = E(Al) = (10 × 10⁶) psi.

But modulus of elasticity of the composite perpendicular to the fibers is given in the expression.

[1 ÷ E(perpendicular)]

= [0.2 ÷ E(fibres)] + [0.8 ÷ E(Al)]

[1 ÷ E(perpendicular)]

= [0.2 ÷ (55 × 10⁶)] + [0.8 ÷ (10 × 10⁶)]

= (3.636 × 10⁻⁹) + (8.00 × 10⁻⁸)

= (8.3636 × 10⁻⁸)

E(perpendicular) = 1 ÷ (8.3636 × 10⁻⁸)

= 11,961,722.5 psi = (11.96 × 10⁶) psi

= (12 × 10⁶) psi

Hope this Helps!!!

6 0
3 years ago
Which descriptions best fit the labels? X: kinetic energy Y: potential energy X: potential energy Y: kinetic energy X: mechanica
GalinKa [24]

its the 3rd option!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

4 0
3 years ago
Read 2 more answers
A uniform plank of mass 10kg and length 10m rests on two supports, A and B as shown. A boy of weight 500N stands at a distance o
kifflom [539]

Answer:

U² = 142.86 N

U¹ = 357.14 N

Explanation:

Taking summation of the moment about point A, we get the following equilibrium equation: (taking clockwise direction as positive)

W(2\ m) - U^2(7\ m) = 0

where,

W = weight of boy = 500 N

U² = reaction ay B = ?

Therefore,

(500\ N)(2\ m)-(U^2)(7\ m)=0\\U^2=\frac{1000\ Nm}{7\ m}\\

<u>U² = 142.86 N</u>

Now, taking summation of forces on the plank. Taking upward direction as positive, for equilibrium position:

W-U^1-U^2=0\\500\ N - 142.86\ N = U^1\\

<u>U¹ = 357.14 N</u>

3 0
3 years ago
How much force is needed to accelerate a toy car of mass 8 kg at 2m/s
atroni [7]

Answer:

4N

Explanation:

f = ma

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4 0
3 years ago
Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2
Tatiana [17]
You can reason it out like this:

-- The car starts from rest, and goes 8 m/s faster every second.

-- After 30 seconds, it's going (30 x 8) = 240 m/s.

-- Its average speed during that 30 sec is  (1/2) (0 + 240) = 120 m/s

-- Distance covered in 30 sec at an average speed of 120 m/s

                                                                           =  3,600 meters .
___________________________________

The formula that has all of this in it is the formula for
distance covered when accelerating from rest:

       Distance = (1/2) · (acceleration) · (time)²

                       = (1/2) ·      (8 m/s²)     · (30 sec)²

                       =      (4 m/s²)          ·      (900 sec²)

                       =            3600 meters.

_________________________________

When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.

When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.

How does he do that ?

By accelerating at 8 m/s².  That's about 0.82 G  !

He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph ! 

He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with.
3 0
3 years ago
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