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Zarrin [17]
3 years ago
8

A horizontal spring with spring constant 85 n/m extends outward from a wall just above floor level. a 3.5 kg box sliding across

a frictionless floor hits the end of the spring and compresses it 6.5 cm before the spring expands and shoots the box back out. how fast was the box going when it hit the spring?
Physics
1 answer:
Rina8888 [55]3 years ago
8 0

k = spring constant of the spring = 85 N/m

m = mass of the box sliding towards the spring = 3.5 kg

v = speed of box just before colliding with the spring = ?

x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m

the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.

Using conservation of energy

Kinetic energy of spring before collision = spring energy of spring after compression

(0.5) m v² = (0.5) k x²

m v² = k x²

inserting the values

(3.5 kg) v² = (85 N/m) (0.065 m)²

v = 0.32 m/s

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When jumping, a flea accelerates at an astounding 1400 m/s2 , but over only the very short distance of 0.55 mm . If a flea jumps
Brums [2.3K]

Answer:

0.079 m or 79 mm

Explanation:

Using the equation of motion

v = √(2as)

Where v is the velocity

a is acceleration = 1400m/s²

s is the distance = 0.55 mm = 0.00055m

Therefore

= √(2 × 1400m/s² × 0.00055 m) = 1.54 m/s

Therefore; initial velocity = 1.54 m/s

Then we use the equation of motion s = v² / 2g

Take g = 9.8 m/s²

Therefore

= (1.54m/s)² / 19.6 m/s²

= 0.079 m or 79 mm

8 0
4 years ago
Read 2 more answers
Two particles, one with charge − 3.77 μC −3.77 μC and one with charge 4.39 μC, 4.39 μC, are 4.34 cm 4.34 cm apart. What is the m
Fudgin [204]

Answer:

the magnitude of the force that one particle exerts on the other is 79.08 N

Explanation:

given information:

q₁ = 3.77 μC = -3.77 x 10⁻⁶ C

q₂ = 4.39 μC = 4.39 x 10⁻⁶ C

r = 4.34 cm = 4.34 x 10⁻² m

What is the magnitude of the force that one particle exerts on the other?

lFl = kq₁q₂/r²

   = (9 x 10⁹) (3.77 x 10⁻⁶) (4.39 x 10⁻⁶)/(4.34 x 10⁻²)²

   = 79.08 N

8 0
4 years ago
when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is remo
Lera25 [3.4K]

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

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

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             \frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1 ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          \frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2}  = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 =   \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         -\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123

                         

                          q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\

 

6 0
4 years ago
How does Newton's law work? Give an example of the formula in a word problem.
Black_prince [1.1K]
Newtons Law works with friction, gravity, and falls.

Also, the formula would be: F = ma

Force = acceleration * mass

Example: The force on a bike is 30 Newtons. The mass of the bike is 6 kg. What is the acceleration of the bike?
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