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Nookie1986 [14]
2 years ago
3

A 0.180 kilogram cart traveling at 0.80 meter per second to the right collides with a 0.100 -kilogram cart initially at rest . T

he carts lock together upon collision Calculate the final velocity of the carts .
Physics
1 answer:
Artist 52 [7]2 years ago
6 0

The final velocity of the two carts is 0.51 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum: in fact, the total momentum of the two carts must be conserved before and after the collision, in absence of external forces. Thus, we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 0.180 kg is the mass of the first cart

u_1 = 0.80 m/s is the initial velocity of the first cart

m_2 = 0.100 kg is the mass of the second cart

u_2 = 0 is the initial velocity of the second cart (at rest)

v is the final combined velocity of the two carts

Re-arranging the equation and substituting the values, we find the  final velocity of the two carts after the collision:  

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(0.180)(0.80)+0}{0.180+0.100}=0.51 m/s

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

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A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w
Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

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1 year ago
The image shows a pendulum that is released from rest at point A. Shari tells her friend that no energy transformation occurs as
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Is  D    the  right  answer
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2 years ago
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A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri
vekshin1

Electric charge on the plastic cube: 1.3\cdot 10^{-7}C

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

V(r)=\frac{kQ}{r}

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Here we have:

V = 200,000 V on the surface of the sphere, so at r = 12.0 cm

We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at

r' = 12.0 + 2.0 = 14.0 cm

Since the voltage is inversely proportional to r, we can use:

Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

qV' = \frac{1}{2}mv^2

where:

q is the charge on the plastic cube

V' is the potential at the location of the cube

m = 5.0 g = 0.005 kg is the mass of the cube

v = 3.0 m/s is the final speed of the cube

Solving for q, we find the charge on the cube:

q=\frac{mv^2}{2V'}=\frac{(0.005)(3.0)^2}{2(171,429)}=1.3\cdot 10^{-7}C

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7 0
2 years ago
A 12.0-kg shell is launched at an angle of 55.0 ∘ above the horizontal with an initial speed of 150 m/s. when it is at its highe
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7 0
2 years ago
An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser
s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

V_L=V_oe^{-\frac{Rt}{L}}

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s

hence, after 1.6s the inductor will have a potential difference of 24V

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2 years ago
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