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3 years ago

7

I love you Please answer my question :-)

2 answers:

Novay_Z [31]3 years ago

6 0

Answer:

A- Astronomical body

C- Galaxy

D- Comet

B- Moon

Hope this helps you! Have a great day!

Vladimir [108]3 years ago

4 0

Answer:

1. A

2. C

3. D

4. B

Explanation:

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Physics Help Please!!!

liq [111]

The answer is the third graph

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3 years ago

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Kate is riding on a train and notices that the wildflowers by the side of the tracks seem to be moving by much faster than the m

madreJ [45]

Answer: motion parallax

Explanation:

Motion parallax refers to a form of depth perception whereby objects that are closer to an individual appears to move at a faster speed than the objects that are far.

Therefore, Kate is riding on a train and notices that the wildflowers by the side of the tracks seem to be moving by much faster than the mountains in the distance is an example of motion parallax.

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3 years ago

A wrench is used to tighten a spark plug. If the wrench handle is 0.20m and a force of 20N is applied, what is the torque applie

Zanzabum

Torque [Nm] = Force [N] x Force arm [m]

C=F*b=20*0.20=4 Nm

The correct answer is C

4 0

3 years ago

What is the slope of the line plotted below? (-2,-1) (4,2) A. 1 B. 2 C. -0.5 D. 0.5

Dmitry [639]

Answer: D. 0.5

Explanation:

The slope formula is y2-y1/x2-x1.

2-(-1)/4-(-2) = 2+1/4+2

2+1/4+2 = 3/6 = 1/2

1/2 = 0.5

The slope is 1/2, or 0.5.

8 0

3 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

so

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

3 years ago