What is the acceleration of a 50 kg object pushed with a net force of 500 newtons?

The diagram shows a swinging pendulum.

pashok25 [27]

Answer:

A. The mechanical energy transforms to thermal energy as the pendulum slows and eventually stops moving.

Explanation:

took the quiz on edge

8 0

3 years ago

Read 2 more answers

Which year was pluto no longer considered a planet?.

katrin2010 [14]

2006, i hope this helps

6 0

1 year ago

Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = $\frac{1}{f}$

where

f = focal length

Thus

f = $\frac{1}{P}$

f = $\frac{1}{2}$ = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}$

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}$

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0

3 years ago

A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that

astraxan [27]

Answer:

A) $k=34867.3384\ N.m^{-1}$

B) $\omega'\approx84\ Hz$

Explanation:

Given:

mass of car, $m=1380\ kg$

A)

frequency of spring oscillation, $f=1.6\ Hz$

We knkow the formula for spring oscillation frequency:

$\omega=2\pi.f$

$\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f$

$\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6$

$k_{eq}=139469.3537\ N.m^{-1}$

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

<u>So, the stiffness of each spring is (as they are identical):</u>

$k=\frac{k_{eq}}{4}$

$k=\frac{139469.3537}{4}$

$k=34867.3384\ N.m^{-1}$

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

$\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }$

$\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }$

$\omega'\approx84\ Hz$

7 0

3 years ago

A block oscillating up and down on a spring comes to the top of its path, where it is momentarily at rest. At the instant it is

True [87]

Opposite to the direction of the velocity which led it to its current position.

Explanation:

The direction of momentum when a vertically oscillating block comes to the rest momentarily will be opposite to the direction of the velocity that it has just followed to reach reach its current position.

The direction of change in momentum at the bottom will be upwards and at the top will be downwards.

The change in momentum is mathematically defined as:

$\Delta P=m.v_f-m.v_i$

where:

$m=$ mass of the block

$v_f=$ final velocity of the block

$v_i=$ initial velocity of the block

When the block comes to rest it is due to the result of continuously decreasing velocity.

7 0

3 years ago