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Assoli18 [71]
3 years ago
12

Blocks A (5.00 kg) and B (10.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is mov

ing toward it at 4.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. What is the maximum energy stored in the spring bumpers during the collision process
Physics
1 answer:
salantis [7]3 years ago
7 0

Answer: 1.33m/s: 13.326j

Explanation:

M1U1+M2U2=V(M1+M2)

M1=5kg, M2=10kg

U1=4.00m/s, U2=0

5*4+10*0=V(10+5)

20+0=15V

20=15V

V=20/15

V=1.333m/s

K. E=1/2MV^2

K. E= 1/2*15*1.332^2

K. E= 1/2*15*1.77689

K. E=26.65335/2

K. E=13.326j

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Actual plate movements can be made les frequent.
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If a fish experiences an a of 9.8m/s2 over a period of 1.84s, what was the 4v?
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A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe
notsponge [240]

Answer:

Q = \frac{0.068}{E}

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field

while weight of the ball is vertically downwards

so here we have

QE = F_x

mg = F_y

since string makes 30 degree angle with the vertical so we will have

tan\theta = \frac{F_x}{F_y}

tan30 = \frac{QE}{mg}

Q = \frac{mg}{E}tan30

Q = \frac{0.012\times 9.81}{E} tan30

Q = \frac{0.068}{E}

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5 0
3 years ago
in the derivation of the time period of a pendulum in electric field when considering the fbd of bob to find the g effective why
Neko [114]

Answer:

we learned that an object that is vibrating is acted upon by a restoring force. The restoring force causes the vibrating object to slow down as it moves away from the equilibrium position and to speed up as it approaches the equilibrium position. It is this restoring force that is responsible for the vibration. So what forces act upon a pendulum bob? And what is the restoring force for a pendulum? There are two dominant forces acting upon a pendulum bob at all times during the course of its motion. There is the force of gravity that acts downward upon the bob. It results from the Earth's mass attracting the mass of the bob. And there is a tension force acting upward and towards the pivot point of the pendulum. The tension force results from the string pulling upon the bob of the pendulum. In our discussion, we will ignore the influence of air resistance - a third force that always opposes the motion of the bob as it swings to and fro. The air resistance force is relatively weak compared to the two dominant forces.

The gravity force is highly predictable; it is always in the same direction (down) and always of the same magnitude - mass*9.8 N/kg. The tension force is considerably less predictable. Both its direction and its magnitude change as the bob swings to and fro. The direction of the tension force is always towards the pivot point. So as the bob swings to the left of its equilibrium position, the tension force is at an angle - directed upwards and to the right. And as the bob swings to the right of its equilibrium position, the tension is directed upwards and to the left. The diagram below depicts the direction of these two forces at five different positions over the course of the pendulum's path.

that's what I know so far

8 0
3 years ago
What is the frequency of this wave?
Reptile [31]
It’s 4 because a coiled springs is closely spaced then widen
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