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Travka [436]
3 years ago
12

You hold glider AA of mass 0.125 kgkg and glider BB of mass 0.375 kgkg at rest on an air track with a compressed spring of negli

gible 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 AA is moving to the right at 0.600 m/sm/s. What is the velocity (magnitude and direction) of glider BB at this time
Physics
1 answer:
andrew11 [14]3 years ago
3 0

Answer:

The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction  

Explanation:

Inelastic Collision

Given data

mass of glider A m1= 0.125kg

initial velocity u1=0

final velocity v1= 0.600 m/s

mass of glider B m2= 0.375kg

initial velocity u2=0

final velocity v2=?

We know that the expression for the conservation of momentum is given as

m1u1+m2u2=m1v1+m2v2

since u1=u2=u=0m/s

u(m1+m2)=m1v1+m2v2

substituting we have

0(0.125+0.0375)=0.125*0.6+0.375*v2

0=0.075+0.375v2

0.375v2=-0.075

v2=-0.075/0.375

v2=-0.2m/s

 The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction      

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A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude
miskamm [114]

Answer:

Explanation:

Given

Weight of person W=652\ N

At highest point Magnitude of the normal force F=585\ N

net force at highest point

F_{net}=W+F_c

where F_c= centripetal force

F_c=\dfrac{mv^2}{r}

F_{net}=F_{n}= Normal Force

585=652+F_c

F_c=-67\ N

Negative sign shows force is in upward direction

At bottom point centripetal force is towards the bottom

F_n=F_c+W

F_n=652+67

F_n=719\ N  

8 0
4 years ago
A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car
kogti [31]

Answer:

t_1 = \frac{v_i}{a_i}

t_2 = \frac{v_i}{a_i}

Δd = v_it_1 = v_i^2/a_i

Explanation:

As v(t) = v_i + at, when the car is making full stop, v(t_1) = 0 . a = -a_i . Therefore,

0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}

Apply the same formula above, with v(t_2) = v_i and a = a_i, and the car is starting from 0 speed,  we have

v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}

As s(t) = vt + \frac{at^2}{2}. After t = t_1 + t_2, the car would have traveled a distance of

s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}

Hence s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}

As t_1 = t_2 we can simplify s(t) = v_it_1

After t time, the train would have traveled a distance of s(t) = v_i(t_1 + t_2) = 2v_it_1

Therefore, Δd would be 2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i

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3 years ago
Which of the following are known to exist on the moon? select all that apply.
denis-greek [22]
B. ice and d basalt............
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Read 2 more answers
a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?
Verdich [7]

The distance between the two charges is 0.3 m

Explanation:

The electrostatic force between two charged objects is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the charges of the two objects

r is the separation between the two charges

In this problem, we are given the following:

q_1 = 8\cdot 10^{-5} C

q_2 = 3\cdot 10^{-5} C

F=2.4\cdot 10^2 N

Therefore, we can rearrange the equation to solve for r, the distance between the two charges:

r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(8\cdot 10^{-5})(3\cdot 10^{-5})}{2.4\cdot 10^2}}=0.3 m

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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Which statement is supported by the information in the diagram and table below?
Natasha_Volkova [10]

Answer:

a

Explanation:

a

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