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Murljashka [212]
3 years ago
12

Masses m and 2m approach each other at the same speed v and collide head-on. Find the final speed of mass 2m, if mass rebounds a

t speed 2v. g
Physics
1 answer:
antoniya [11.8K]3 years ago
6 0

Answer:

<em>The rebound speed of the mass 2m is v/2</em>

Explanation:

I will designate the two masses as body A and body B.

mass of body A = m

mass of body B = 2m

velocity of body A = v

velocity of body B = -v    since they both move in opposite direction

final speed of mass A = 2v

final speed of body B = ?

The equation of conservation of momentum for this system is

mv - 2mv = -2mv + x

where x is the final momentum of the mass B

x = mv - 2mv + 2mv

x = mv

to get the speed, we divide the momentum by the mass of mass B

x/2m = v = mv/2m

speed of mass B = <em>v/2</em>

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Calculate the current through a 3.0 ω resistor with a voltage of 9.0 v across it.
gulaghasi [49]

Answer: 3 A

Explanation:

According to<u> Ohm's law</u>:  

V=R.I

Where:

V=9 V is the voltage

R=3\Omega is the resistance of the resistor

I is the electric current (the value we want to find)

Isolating I:

I=\frac{V}{R}

I=\frac{9 V}{3\Omega}

Finally:

I=3 A

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3 years ago
There is a puncture in the inner tor of your bicycle. Give a method to detect the position of the puncture
Pani-rosa [81]

Answer:

Explanation:

The first method to engage is to listen to where the sound of air in the inner Tor escaping originated and look to see if u can find it. You can then feel the escape air with your hand.

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7 0
3 years ago
8. A turtle crawls along a straight line, which we will call the x-axis with the positive direction to the right. The equation f
pychu [463]

(a) The turtle's initial velocity is 4 m/s, initial position of the turtle is 5 cm, and initial acceleration is -1.25 m/s².

(b) The time when the velocity of the turtle is zero is 3.2 s.

(c) The time taken for the turtle to return to its starting point is 6.4 s.

(d) The time taken for the turtle to travel 30 cm is 0.08 s.

<h3>Initial velocity of the turtle</h3>

The initial velocity of the turtle is calculated as follows;

v = \frac{dx}{dt} \\\\x = 5 + 4t -0.625t^2\\\\v(t) = 4 - 1.25t\\\\v(0) = 4-0\\\\v(0) = 4 \ m/s

<h3>Initial acceleration of the turtle</h3>

The initial acceleration of the turtle is calculated as follows;

a = \frac{dv}{dt} \\\\v(t) = 4 - 1.25t\\\\a = -1.25\ m/s^2

<h3>Initial position of the turtle</h3>

x(t) = 5 + 4t - 0.625t²

x(0) = 5 cm

<h3>Time when the velocity becomes zero</h3>

v(t) = 4 - 1.25t

0 = 4 - 1.25t

1.25t = 4

t = 4/1.25

t = 3.2 s

<h3>Time taken to return to starting point</h3>

The total distance traveled is calculated as follows

v² = u² + 2ad

0 = (4)² + 2(-1.25)d

0 = 16 - 2.5d

2.5d = 16

d = 16/2.5

d = 6.4 m

Time to travel the given distance;

d = ut + ¹/₂at²

6.4 = (4)t + ¹/₂(-1.25)t²

6.4 = 4t - 0.625t²

0.625t² - 4t + 6.4 = 0

solve the quadratic equation using formula method;

t = 3.2 s

The time travel the distance two times, = 2 x 3.2 s = 6.4 s

<h3>Time taken for the turtle to travel 30 cm</h3>

d = ut + ¹/₂at²

0.3 = (4)t + ¹/₂(-1.25)t²

0.3 = 4t - 0.625t²

0.625t² - 4t + 0.3 = 0

solve the quadratic equation using formula method;

t = 0.08 s

Learn more about velocity here: brainly.com/question/6504879

7 0
2 years ago
When a cold air mass and a warm air mass meet but are at a standstill, the boundary is called.
Sphinxa [80]

Answer:

Ur answer is Stationary Front

Explanation:

Stationary Front is when a cold air mass and a warm air mass but are at a standstill the boundary is called Stationary Front.

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