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3 years ago

7

What is average velocity?

2 answers:

Ber [7]3 years ago

6 0

The average speed of an object is defined as the distance traveled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time. According To a website on google

Alex17521 [72]3 years ago

5 0

Average Velocity<span>, General. The </span>average<span> speed of an object is defined as the distance traveled divided by the time elapsed. </span>Velocity<span> is a vector quantity, and </span>average velocity<span> can be defined as the displacement divided by the time.</span>

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

a)

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

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3 years ago

What is the name of the process in which oceanic crust moves apart at mid-ocean ridges due to convention currents in the mantle?

Katena32 [7]

The answer is D seafloor spreading

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3 years ago

Ngle of a block is 45 degrees. What is the refractive index

lyudmila [28]

1.6 ??? I hope I’m right

3 0

3 years ago

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

5 0

3 years ago

Who is wwe universal champion

mina [271]

Answer:

Roman reigns

Explanation:

is the wwe universal hampion

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3 years ago