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3 years ago

What happens to the object if the line crosses the x-axis from the positive portion of a velocity versus time diagram?

Physics

1 answer:

stealth61 [152]3 years ago

7 0

Answer:

A velocity time graph shows the change of velocity of an object with respect ot time. If the slope of the graph is increasing in the postive region, it means that the velocity is changing, if the slope is decreasing, it means the the velocity is decreasing, but the object is moving in the same direction (positve direction).

If this slope intersects the graph at x-axis, it means that the body has 0 velocity and has become still. After that, if the line enters in the negative region, it means that its velocity is started to increases again, but the body is movinging in the opposite direction (negative direction)

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A 1200 kg car traveling north at 10 m/s is rear-ended by a 2000 kg truck traveling at 30 m/s. What is the total momentum before

tino4ka555 [31]

Answer:

The total momentum before and after collision is 72000 kg-m/s.

Explanation:

Given that,

Mass of car = 1200 kg

Velocity of car = 10 m/s

Mass of truck = 2000 kg

Velocity of truck = 30 m/s

Using conservation of momentum

The total momentum before the collision is equal to the total momentum after collision.

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})V$

Where, $m_{1}$ =mass of car

$v_{1}$ =velocity of car

$m_{1}$ =mass of truck

$v_{1}$ =velocity of truck

Put the value into the formula

$1200\times10+2000\times30=(1200+2000)V$

$V=\dfrac{1200\times10+2000\times30}{(1200+2000)}$

$V = 22.5\ m/s$

Now, The total momentum before collision is

$P=m_{1}v_{1}+m_{2}v_{2}$

$P=1200\times10+2000\times30$

$P=72000\ kg-m/s$

The total momentum after collision is

$P=(m_{1}+m_{2})v_{2}$

$P=(1200+2000)\times22.5$

$P= 72000 kg-m/s$

Hence, The total momentum before and after collision is 72000 kg-m/s.

4 0

3 years ago

The Bohr radius a0 is the most probable distance between the proton and the electron in the Hydrogen atom, when the Hydrogen ato

katen-ka-za [31]

Answer:

The electric force is $F = 11.9 *10^{-9} \ N$

Explanation:

From the question we are told that

The Bohr radius at ground state is $a_o = 0.529 A = 0.529 ^10^{-10} \ m$

The values of the distance between the proton and an electron $z = 2.63a_o$

The electric force is mathematically represented as

$F = \frac{k * n * p }{r^2}$

Where n and p are charges on a single electron and on a single proton which is mathematically represented as

$n = p = 1.60 * 10^{-19} \ C$

and k is the coulomb's constant with a value

$k =9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F = \frac{9*10^{9} * [(1.60*10^{-19} ]^2)}{(2.63 * 0.529 * 10^{-10})^2}$

$F = 11.9 *10^{-9} \ N$

3 0

3 years ago

At what speed do a bicycle and its rider, with a combined mass of 100 kg, have the same momentum as a 1400 kg car traveling at 2

likoan [24]

100kg x bicycle speed = 1400 X 2

bicycle speed = 2800/ 100
bicycle speed = 28 m/s

3 0

4 years ago

In which ancient civilization did the sport of cucj originate

Burka [1]

It originated in China.

3 0

3 years ago

If the car speeds up at a steady 1.6 m/s2 , how long after starting is the magnitude of its centripetal acceleration equal to th

Rufina [12.5K]

Based on internet sources, the basic formulas are: v^2/r = (at)^2/r = a ==> at^2 = r ==> t = sqrt(r/a).

Assuming the missing units are mutually compatible, as in the following example, they don't need to be known. 
Acceleration = 1.6 cramwells/s^2 
Radius = 150 cramwells 
t = sqrt(150/1.6) = 9.68 s.

I hope this helps.

8 0

3 years ago