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exis [7]
3 years ago
8

Find change of velocity for: 0-1.7 s 1.7-6s

Physics
1 answer:
tekilochka [14]3 years ago
3 0
The integral of acceleration is velocity. The area under the curve is an integral. You can see the relation here. So just take the area under those time intervals. I’m assuming 1.7 is where it crosses the x-axis

For the first interval:
Base = 1.7
Height = -2
[1.7*(-2)]/2 = -1.7 cm/s

For the second interval:
There are two triangles and a rectangle here.
Base #1 = 0.3
Height #1 = 1

Base #2 = 1
Height #2 = 1

Length = 3
Width = 1

Apply the area formulas to get an answer of 3.65 cm/s
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Explain how birthrates and death rates influence the size of population.
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Having a high birth rate and a slow death rate would lead to an increase of population but a low birth rate and a high death rate would lead to the decrease of a population.
4 0
3 years ago
If a wave has a frequency of 5 Hz, what is its period?
tino4ka555 [31]

Answer: period = 0.2 s

Explanation:

5Hz=0.2s

6 0
3 years ago
A 98-kg fullback is running along at 8.6 m / s when a 76-kg defensive back running in the same direction at 9.8 m / s jumps on h
algol13
The total momentum before and after the collision must be conserved.

The total momentum before the collision is:
p_i = m_1 v_1 + m_2 v_2
where m1 and m2 are the masses of the two players, and v_1 and v_2 their initial velocities. Both are considered with positive sign, because the two players are running toward the same direction.

The final momentum is instead
p_f = (m_1+m_2)v_f
because now the two players are moving together with a total mass of (m1+m2) and final speed vf.

By requiring that the momentum is conserved
p_i=p_f
we  can calculate vf, the post-collision speed:
m_1 v_1 + m_2 v_2 = (m_1+m_2)v_f
v_f =  \frac{m_1 v_1 + m_2 v_2}{m_1 +m_2}= \frac{(98 kg)(8.6 m/s)+(76 kg)(9.8m/s)}{98 kg+76 kg}=9.1 m/s
and the direction is the same as the direction of the players before the collision.
6 0
3 years ago
Physicists and engineers from around the world have come together to build the largest accelerator in the world, the Large Hadro
elena-14-01-66 [18.8K]

Solution :

Energy of photon, E = 6.7 eV

                              E = $6.7 \times 1.602 \times 10^{-7}$ joule

Kinetic energy, $K.E. =\frac{1}{2} mv^2 = 1.602 \times 6.7 \times 10^{-7}$

$v^2=\frac{2 \times 1.602 \times 6.7 \times 10^{-7}}{1.6726 \times 10^{-27}}$

   $=12.834 \times 10^{-20}$

Kinetic energy at high speeds

$(r-1)\times mc^2 = 6.7 \ eV$

$(r-1)=\frac{6.7 \times 1.602 \times 10^{-7}}{1.6726 \times 10^{-27} \times 9 \times 10^{16}}$

r - 1 = 7130

r = 7130 + 1

r  = 7131

$\frac{1}{\sqrt{1-\frac{v^2}{C^2}}}=7131$

$1-\frac{v^2}{C^2} = \left(\frac{1}{7131}\right)^2$

$v^2=C^2\left[1-\left(\frac{1}{7131}\right)^2\right]$

$v=0.99999999017C$

Δ = 1 - 0.99999999017

   = 0.00000000933

Relative mass, $m_{rel}=r.m$

                                $=7131 \times 1.6728 \times 10^{-27}$

                               $=1.1927 \times 10^{-23}$ kg

                                 

6 0
3 years ago
Nuclide X has a higher rate of decay than nuclide Y. Based on this information, which of the following statements must be true?
marshall27 [118]

Answer: Answer down below.

Explanation:

8 0
2 years ago
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