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3 years ago

12

A weightlifter lifts a set of 1250kg weights a vertical distance of 2m weight lifting contest. what potential energy do the weig

hts now possess

1 answer:

Juliette [100K]3 years ago

4 0

Answer:

24525 J

Explanation:

Using Mgh as the calculation for potential energy

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A boy pushes a 5 kg box across the floor with a force of 25 N. There is a 5 N frictional force opposing him. What is the net for

11Alexandr11 [23.1K]

Answer:net force:20N, accel. :4m/s2

Explanation:

8 0

3 years ago

A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s

____ [38]

Answer:

Speed of both blocks after collision is 2 m/s

Explanation:

It is given that,

Mass of both blocks, m₁ = m₂ = 1 kg

Velocity of first block, u₁ = 3 m/s

Velocity of other block, u₂ = 1 m/s

Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{1\ kg\times 3\ m/s+1\ kg\times 1\ m/s}{2\ kg}$

v = 2 m/s

Hence, their speed after collision is 2 m/s.

7 0

3 years ago

URGENT: Testing shows that a sample of wood from an artifact contains 50% of the

Murrr4er [49]

Answer:

Im so so sorry but I do now know how to do this

Explanation:

4 0

3 years ago

To construct an oscillating LC system, you can choose from a 11 mH inductor, a 6.0 μF capacitor, and a 4.2 μF capacitor. What ar

Free_Kalibri [48]

Answer:

a. 475.14 Hz

b. 1959 Hz

c. 2341.53 Hz , 3053.34 Hz

Explanation:

$f = \frac{1}{2\pi*\sqrt{C*L}}$

a. smallest use the capacitive 4.2 uF + 6.0 uF = 10.2uF replacing:

$f = \frac{1}{2\pi*\sqrt{C*L}}f=\frac{1}{2\pi*\sqrt{10.2uF*11mH}}$

$f = 475.14 Hz$

b. second smallest use the capacitive 6 uF so:

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{6uF*11mH}}$

$f = 1959Hz$

c. second largest and largest oscillation first combination so:

Use 4.2 uF

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{4.2uF*11mH}}$

$f = 2341.53 Hz$

And finally largest oscillation cap in serie so:

$C=\frac{c_1*c_2}{c_1+c_2}=\frac{4.2uF*6.0uf}{4.2uf+6.0uF}$

$C=2.47 uF$

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{2.47uF*11mH}}$

$f = 3053.34 Hz$

5 0

3 years ago

Hello....I need help with this question.

musickatia [10]

Answer:

3.2 m/s²

Explanation:

Acceleration can be calculated as:

v = u + at (where v is final velocity, u is initial velocity, a is acceleration and t is time)

25 m/s = 9 m/s + a(5 s) (a is unknown)

16 m/s = a(5 s)

a = 3.2 m/s²

We assume that this is a uniform acceleration (meaning that the velocity increases at an equal rate for those 5 seconds).

6 0

3 years ago

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