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pav-90 [236]
2 years ago
9

Convert 3.8 kg to Lbs

Physics
2 answers:
JulijaS [17]2 years ago
8 0
I believe it’s 8.3 to be precise
Ray Of Light [21]2 years ago
3 0

Answer:

8.37757

Explanation:

may i have brainliest

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What do an earthquake wave and a light wave have in common?
rusak2 [61]
It’s c.) I think so at least
5 0
3 years ago
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Two cars having different weights are traveling on a level surface at different constant velocities. Within the same time interv
GuDViN [60]

I think you're saying that once you start pushing on the cars, you want to be able to stop each one in the same time. 

This is sneaky.  At first, I thought it must be both 'c' and 'd'.  But it's not
kinetic energy, for reasons I'm not ambitious enough to go into.
(And besides, there's no great honor awarded around here for explaining
why any given choice is NOT the answer.)

The answer is momentum.

Momentum is (mass x speed).  Change in momentum is (force x time).

No matter the weight (mass) or speed of the car, the one with the greater
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4 0
3 years ago
A 150kg person stands on a compression spring with spring constant 10, 000 N/m and norminal.Length of 0.50.What is the togal len
Eva8 [605]

Answer:

Total length of spring 0.647 m

Explanation:

We have given mass of the person m = 150 kg

Acceleration due to gravity g=9.8m/sec^2

Spring constant k = 10000 N/m

Nominal length of spring = 0.50

According to hook's law

mg=kx

150\times 9.8=10000\times x

x = 0.147 m

So total length of spring = 0.50+0.147 = 0.647 m

4 0
3 years ago
What properties of the wave define why it is found within this area of the spectrum?
MatroZZZ [7]

Answer:

Explanation:

Speed and medium are properties of a wave which is in common within an area of the spectrum of visible light. i.e;

Their medium of propagation

They both travel at the same speed ( speed of light )

The properties mentioned above are properties that define that wave is found within an area spectrum for visible light.

Wish I Could Help You!

7 0
2 years ago
Baseball player a bunts the ball by hitting it in such a way that it acquires an initial velocity of 2.4 m/s parallel to the gro
LUCKY_DIMON [66]

Let \mathbf r_A denote the position vector of the ball hit by player A. Then this vector has components

\begin{cases}r_{Ax}=\left(2.4\,\frac{\mathrm m}{\mathrm s}\right)t\\r_{Ay}=1.2\,\mathrm m-\frac12gt^2\end{cases}

where g=9.8\,\dfrac{\mathrm m}{\mathrm s^2} is the magnitude of the acceleration due to gravity. Use the vertical component r_{Ay} to find the time at which ball A reaches the ground:

1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s

The horizontal position of the ball after 0.49 seconds is

\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector \mathbf r_B of the ball hit by player B has

\begin{cases}r_{Bx}=v_0t\\r_{By}=1.6\,\mathrm m-\frac12gt^2\end{cases}

Again, we solve for the time it takes the ball to reach the ground:

1.6\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.57\,\mathrm s

After this time, we expect a horizontal displacement of 12 meters, so that v_0 satisfies

v_0(0.57\,\mathrm s)=12\,\mathrm m

\implies v_0=21\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
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