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Monica [59]
3 years ago
6

a force of 25.0 newtons is applied so as to move a 5.0 kg mass a distance of 20.0 meters. How much work was done? with work plea

se
Physics
1 answer:
miv72 [106K]3 years ago
4 0
First we need to know the equation of work
work=force times distance
w= 25 N times 20 meters
w= 500 joules
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3 0
3 years ago
SP1b.
nata0808 [166]

Answer:

2 m/s^2, west

Explanation:

Vf=final velcoity

Vi=initial velocity

t=timw

a =  \frac{vf - vi}{t}

=

\frac{15 - 25}{5}

= - 2 m/s^2

The - changes direction and makes it opposite

2 m/s, west

3 0
3 years ago
A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len
weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}

\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}

\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

6 0
3 years ago
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