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bazaltina [42]
4 years ago
11

You hold glider A of mass 0.125 kg and glider B of mass 0.375 kg at rest on an air track with a compressed spring of negligible

mass between them. When you release the gliders, the spring pushes them apart. Once the gliders are no longer in contact with the spring, glider A is moving to the right at 0.600 m/s. What is the velocity (magnitude and direction) of glider B at this time?
Glider A moving to the right at 0.600 m/s then collides head-on with a third glider C of mass 0.750 kg that is moving to the left at 0.400 m/s. After this collision, glider C is moving to the left at 0.150 m/s. What is the velocity (magnitude and direction) of glider A after this collision?

Physics
1 answer:
Yuliya22 [10]4 years ago
8 0

Answer: a) final velocity of glider B = 0.20m/s ( in the left direction)

b) final velocity of glider A = 0.9m/s ( in the left direction)

Explanation: all shown in the attachment.

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Exercice 1
Mariana [72]

Answer:

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1.3 Calculate the molar concentration of these fructose solutions

1.4

6 0
3 years ago
Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repul
Gnom [1K]

Answer:

t=2.5\times 10^{-14}\ s

Explanation:

We know that charge on electron

q=1.6\times 10^{-19}\ C

r= 2 nm

We know that force between two charge given

F=K\dfrac{Q_1Q_2}{r^2}

Now by putting the value

F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}

F=5.67\times 10^{-11}\ N

We know that mass of electron

The mass of electron

m=9.1\times 10^{-31}\ kg

F= m a

a= Acceleration of electron

a= F/m

a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2

a=6.2\times 10^{19} m/s^2

S=ut+\dfrac{1}{2}at^2

initial velocity given that zero ,u=0

20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2

t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}

t=2.5\times 10^{-14}\ s

3 0
4 years ago
Which of the following is most likely to be an observation made by a physiologist
Naya [18.7K]
Do you have the answer choices ?
3 0
4 years ago
A 20 kg crate initially at rest on a horizontal floor requires a 80 N horizontal force to set it in motion. Find the coefficient
e-lub [12.9K]

Answer:

<em>The coefficient of static friction between the crate and the floor is 0.41</em>

Explanation:

<u>Friction Force</u>

When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

Fr=\mu N          [1]

Where \mu is the coefficient of static or kinetics friction and N is the normal force.

If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:

N = W = m.g

The crate of m=20 Kg has a weight of:

W = 20*9.8

W = 196 N

The normal force is also N=196 N

We can find the coefficient of static friction by solving [1] for \mu:

\displaystyle \mu=\frac{Fr}{N}

The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:

\displaystyle \mu=\frac{80}{196}

\mu=0.41

The coefficient of static friction between the crate and the floor is 0.41

7 0
3 years ago
If we increase the force applied to an object and all other factors remain the same that amount of work will
worty [1.4K]
Hello there.

<span>If we increase the force applied to an object and all other factors remain the same that amount of work will

</span><span>C. Increase
</span>
5 0
3 years ago
Read 2 more answers
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