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Katyanochek1 [597]
3 years ago
7

You did 130 J of work lifting a 100 N backpack. How high did you lift the backpack?

Physics
1 answer:
Ronch [10]3 years ago
5 0

Answer:

1.3 m

Explanation:

The work done in lifting the backpack is equal to the change in gravitational potential energy of the backpack, so:

\Delta U=W \Delta h

where

W = 100 N is the weight of the backpack

\Delta h is the change in heigth of the object

In this problem, we know that

\Delta U = 130 J

so we can re-arrange the equation to find the change in height of the backpack:

\Delta h = \frac{\Delta U}{W}=\frac{130 J}{100 N}=1.3 m

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Which water depth had the biggest difference in survival rates for embryos with UV-B protection versus embryos without UV-B prot
sergiy2304 [10]

Answer:

10 cm

Explanation:

water depth had the biggest difference in survival rates for embryos with UV-B protection versus embryos without UV-B protection is 10 cm . the bar graph attached in the shows that the lengths of yellow and brown bars differ the most at 10 cm.

Hence,The biggest difference between the survival rates of UV-B protected and unprotected can be seen at the depth of 10 cm.

6 0
3 years ago
Give reason : the action and reaction do not lead to equilibrium ?
Natalija [7]

the action and reaction do not lead equilibrium if action and reaction force react on different objects

4 0
3 years ago
Consider a series RLC circuit where R = 855 Ω and C = 6.25 μF. However, the inductance L of the inductor is unknown. To find its
sashaice [31]

Answer:

L= 0.059 mH

Explanation:

Given that

R = 855 Ω and C = 6.25 μF

V= 84 V

Frequency

ω = 51900 1/s

We know that

\omega=\sqrt{\dfrac{1}{LC}}

L=Inductance

C=Capacitance

ω =angular Frequency

ω² L C =1

(51900)² x L x 6.25 x 10⁻⁶ = 1

L= 5.99 x 10⁻⁵ H

L= 0.059 mH

6 0
3 years ago
There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in
LekaFEV [45]

Answer:

c = 3.00E108 m/s = 3.00E5 km/s

t = S / v = 3.84E5 / 3.00E5 = 1.28 sec

4 0
3 years ago
Can someone help?
laila [671]

Answer:

i) 21 cm

ii) At infinity behind the lens.

iii) A virtual, upright, enlarged image behind the object

Explanation:

First identify,

object distance (u) = 42 cm (distance between  object and lens, 50 cm - 8 cm)

image distance (v) = 42 cm (distance between  image and lens, 92 cm - 50 cm)

The lens formula,

\frac{1}{v} -\frac{1}{u} =\frac{1}{f}

Then applying the new Cartesian sign convention to it,

\frac{1}{v} +\frac{1}{u} =\frac{1}{f}

Where f is (-), u is (+) and  v is (-) in  all 3  cases. (If not values with signs have to considered, this method that need will not arise)

Substituting values you get,

i) \frac{1}{42} +\frac{1}{42} =\frac{1}{f}\\\frac{2}{42} =\frac{1}{f}

f = 21 cm

ii) u =21 cm, f = 21 cm v = ?

Substituting in same equation\frac{1}{v} =\frac{1}{21} =\frac{1}{21} \\\\\frac{1}{v} = 0\\

  v ⇒ ∞ and image will form behind the lens

iii) Now the object will be within the focal length of the lens. So like in the attachment, a virtual, upright, enlarged image behind the object.

7 0
3 years ago
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