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GREYUIT [131]
4 years ago
13

an 11 kg tool box is floating stationary in an orbiting spacecraft when a 79 kg astronaut, initially at rest, gives the tool box

as apush. if the tool box moves with a velocity of 0.45 m/s to the left, what is the astronauts final velocity
Physics
1 answer:
snow_tiger [21]4 years ago
6 0

The astronaut final velocity is 0.06 m/s to the right

Explanation:

In absence of external forces, the total momentum of the box-astronaut system is conserved.

At the beginnig, the total momentum of the two is zero, since they are at rest:

p_i = 0

While the final momentum, after the astronaut throws the box, is:

p_f=mv+MV

where

m = 11 kg is the mass of the box

M = 79 kg is the mass of the astronaut

v = -0.45 m/s (to the left) is the velocity of the box (we take left as negative direction)

V is the final velocity of the astronaut

The total momentum is conserved, so

p_i = p_f\\0=mv+MV

And solving , we find V:

V=-\frac{mv}{M}=-\frac{(11)(-0.45)}{79}=0.06 m/s

And the positive sign indicates that the direction is to the right.

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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Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and
Lelu [443]

Answer:

The maximum shear stress in shaft AB, T_{ABmax} is 15 MPa

The maximum shear stress in shaft CD,  T_{CDmax} is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by  

J = \pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}

Therefore, we have

J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}

Where:

D_{AB} = Diameter of shaft AB = 30 mm = 0.03 m

r_{AB} = Radius of shaft AB = 15 mm = 0.015 m

∴ J_{AB} = \pi \times \frac{0.03^4}{32} =\pi \times \frac{0.015^4}{2} = 7.95 × 10⁻⁸ m⁴

and

J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}

Where:

D_{CD} = Diameter of shaft CD = 36 mm = 0.036 m

r_{CD} = Radius of shaft CD = 18 mm = 0.018 m

Therefore,

J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2} = 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= 1.58 \times \frac{2\pi }{360} rad = 2.76 × 10⁻² rad.

That is \phi_r = 2.76 × 10⁻² rad.

However  \phi_r =  \phi_{C/D} -  \phi_{B/A}  

Where:

\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G} and

\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}

T_{AB} and T_{CD}= Torque on shaft AB and CD respectively

T_{AB}  = Required

T_{CD}= 500 N·m

L_{AB} and L_{CD} = Length of shafts AB an CD respectively

L_{AB}  = 600 mm = 0.6 m

L_{CD} = 900 mm = 0.9 m

G = Shear modulus of the material = 77.2 GPa

Therefore;

\phi_r =  \phi_{C/D} -  \phi_{B/A}  =\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}

2.76 × 10⁻² rad =\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}

=\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}

Therefore;

T_{AB} =  79.54 N.m

Where T = T_{AB} + T_{CD} =

Therefore T_{CD total } = 500 - 79.54 = 420.46 N·m

τ_{max} = \frac{T\times R}{J}

\tau_{ABmax} = \frac{T_{AB}\times R_{AB}}{J_{AB}} =  \frac{79.54\times 0.015}{7.95\times 10^{-8}} = 15 MPa

\tau_{CDmax} = \frac{T_{CD}\times R_{CD}}{J_{CD}} = \frac{420.46\times 0.018}{1.65\times 10^{-7}} = 45.9 MPa

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To balance a chemical equation we use______. <br> -coefficients <br> -subscripts
Vitek1552 [10]

Answer:

Coefficients

Explanation:

  • Coefficients are used to balance chemical equations.
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  • The conservation of mass is achieved in chemical equations is achieved through balancing chemical equations.
  • <em><u>Balancing chemical equations is a try and error of putting appropriate coefficients to the reactants or products to ensure that the number of atoms of each element are equal on the side of the reactants and that of products. </u></em>
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4 years ago
car company wants to build a wind- powered car that converts 100 percent of the mechanical energy in the wind to the mechanical
navik [9.2K]

EVEN IF they can build such a machine, it's not too useful.

-- If the wind starts and stops, your car would do the same thing.

-- If the wind isn't blowing at all, your car is going nowhere. (and fast)

-- You could never move along the road faster than the wind is moving along the road.

-- You could never move in the direction towards where the wind is coming from.  This has been proven before, with the technological marvel known as the "sailboat".

Call your broker immediately.  Tell him you do NOT want to buy any stock in CarCompany.

3 0
3 years ago
Read 2 more answers
A small rock is thrown vertically upward with a speed of 27.0 m/s from the edge of the roof of a 21.0-m-tall building. The rock
lbvjy [14]

Answer:

A. 33.77 m/s

B. 6.20 s

Explanation:

Frame of reference:

Gravity g=-9.8 m/s^2; Initial position (roof) y=0; Final Position street y= -21 m

Initial velocity upwards v= 27 m/s

Part A. Using kinematics expression for velocities and distance:

V_{final}^{2}=V_{initial}^{2}+2g(y_{final}-y_{initial})\\V_{f}^{2}=27^{2}-2*9.8(-21-0)=33.77 m/s

Part B. Using Kinematics expression for distance, time and initial velocity

y_{final}=y_{initial}+V_{initial}t+\frac{1}{2} g*t^{2}\\==> -0.5*9.8t^{2}+27t+21=0\\==> t_1 =-0.69 s\\t_2=6.20 s

Since it is a second order equation for time, we solved it with a calculator. We pick the positive solution.

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3 years ago
Which of the following is an example of a chemical change?
bija089 [108]

HELLO THERE

THE ANSWER IS

D. A gas when vinegar and baking soda are mixed

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