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nasty-shy [4]
3 years ago
6

Show that Ns is the same as kgmls.​

Physics
1 answer:
Minchanka [31]3 years ago
3 0

1 N = 1kgm/s²

then,

1Ns = 1kgm/s² * s

1Ns = 1kgm/s

Hence shown.

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Two trolleys are moving in the same direction along a track. Trolley 1 has a momentum of 2 kg m/s and Trolley 2 has a momentum o
motikmotik

Hi there!

Recall the conservation of momentum:

\large\boxed{m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'}}

For this type of inelastic collision:

\large\boxed{m_1v_1 + m_2v_2 = v_f(m_1 + m_2)}}

Thus, the initial momentum equal the final momentum if there are no external forces.

We can begin by writing out this problem:

2 + 6 = v_f(m_1 + m_2)

8 = 2(m_1 + m_2)\\\\m_1 + m_2 = \boxed{4 kg}

8 0
3 years ago
Can someone tell me how to get the answers for 11.a) and b)
faust18 [17]
11. (a)
The instantaneous speed when t=6 sec is the slope of the curve at that point.

Did you draw the red line on the graph ?  That's absolutely the way
to do part-(a), but the line you draw should kiss the curve at the next
point ... the point on the curve where t=6, not the point where t=5.

Draw another line that just touches the curve where t=6.  Then, as well as
you can, measure the slope of that line.  You'll have to do some measuring
and some estimating.  Mark off a section of the line you draw.  Then measure
how much horizontal distance that section covers, and how much vertical distance
it covers.

The slope of the line is  (vertical distance) divided by (horizontal distance).
Its unit is  meters/second. 

11.  (b)
Finding the average speed is much easier than finding the instantaneous speed.
The average is just the slope of a line between the ends of the graph, and you
don't care how the graph twists and turns between the start and end points.

The point at the beginning is      (0 sec,  0 meters) .

The point at the end is    (10 sec,  200 meters) .

Average speed = (distance covered) divided by (time to cover the distance) .

                            =       200 meters                /                    10 seconds

                            =             20 meters/sec       from zero to 10 seconds

3 0
4 years ago
Pls help me with this thing.<br>Will mark brainliest!!!​
IgorLugansk [536]
A) Pulley and Hook
B) bridge
C) Wheel and axle
5 0
2 years ago
A book prone to air resistance is released from rest 300 m
yaroslaw [1]

Answer:

Approximately 73\%.

(Assuming that g = \rm 9.81\; m \cdot s^{-2}.)

Explanation:

The mechanical energy of an object is the sum of its potential energy and its kinetic energy. It will be shown that the exact mass of this object doesn't matter. For ease of calculation, let m(\text{book}) represent the mass of the book.

The initial potential energy of the book is  

\begin{aligned}U(300\; \text{m}) &= m(\text{book}) \cdot g \cdot \Delta h + U(0\; \text{m}) \cr &=(9.81 \times 300) \cdot m(\text{book})\cr &= \left(2.943\times 10^3\right) \cdot m(\text{book})\end{aligned}.

The book was initially at rest when it was released. Hence, its initial kinetic energy would be zero. Hence, the initial mechanical energy of the book-Earth system would be (2.943\times 10^3) \cdot m(\text{book}).

When the book was about to hit the ground, its speed is \rm 40\; m \cdot s^{-1}. Its kinetic energy would be:

\begin{aligned} \text{KE} &= \frac{1}{2} \, m(\text{book}) \cdot v^{2} \cr &= \left(\frac{1}{2} \times 40^2\right)\cdot m(\text{book}) \cr &= \left(8.00\times 10^2\right)\cdot m(\text{book})\end{aligned}.

The question implies that the potential energy of the book near the ground is zero. Hence, the mechanical energy of the system would be \left(8.00\times 10^2\right)\cdot m(\text{book}) when the book was about to hit the ground.

The amount of mechanical energy lost in this process would be equal to:

\begin{aligned}&\left(2.943\times 10^3\right) \cdot m(\text{book}) - \left(8.00\times 10^2\right)\cdot m(\text{book}) \cr &=\left(2.143\times 10^3\right)\cdot m(\text{book})\end{aligned}.

Divide that with the initial mechanical energy of the system to find the percentage change. Note how the mass of the book, m(\text{book}), was eliminated in this process.

\begin{aligned}&\frac{\left(2.143\times 10^3\right)\cdot m(\text{book})}{\left(2.943\times 10^3\right) \cdot m(\text{book})}\times 100\% \cr &= \frac{2.143\times 10^3}{2.943\times 10^3}\times 100\% \cr & \approx 73\%\end{aligned}.

5 0
4 years ago
The diagram below represents an electromagnetic wave. Please Help<br>​
forsale [732]

Answer:

I'm pretty sure it's A

Explanation:

bc the crest is is the point on the wave

8 0
3 years ago
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