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4 years ago

REALLY NEED HELP WITH THIS!!!

Physics

1 answer:

goldenfox [79]4 years ago

7 0

I don't know if this is right, but inheritance?

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After a water had boiled, the temperature of the water decreases by 22degrees. The mass of the water in the kettle is 0.50kg.The

kifflom [539]

<u>Answer</u>

46,200 J

The energy of given out when a substance is losing heat is given by:

H = mcΔθ

Where m is the mass of the substance,

c is the specific heat capacity of the substance and

Δθ is the temperature change.

H = 0.50 × 22 4200

= 46,200 Joules.

4 0

4 years ago

The maximum number of electrons in the second energy level of an atom is ____.

ehidna [41]

Your answer will be 8.

6 0

3 years ago

You drop a 30 g pebble down a well. You hear a splash 2.7 s later. Ignoring air resistance, how deep is the well? Assume g = 9.8

lys-0071 [83]

I will assume here that the well is sufficiently short so that the time the sound takes to come from the bottom of the well to our ear is negligible.

Since the pebble moves by uniformly accelerated motion, the distance it covers is given by
$S= \frac{1}{2}gt^2$
where
$g=9.81 m/s^2$ is the gravitational acceleration
$t=2.7 s$ is the time the pebble takes to reach the bottom of the well

Therefore, the depth of the well is
$S= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2 = 35.7 m \sim 36 m$
and the correct answer is B.

8 0

3 years ago

An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens

Pavlova-9 [17]

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

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

Putting the values

$\frac{1}{v} +\frac{1}{15} =\frac{1}{14}$

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

$\frac{210}{15} \times \frac{D}{f_e}$

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

7 0

3 years ago

You can not exert force on a wall unless

ludmilkaskok [199]

unless...the wall simulaneiously exerts the same amount of force on you.

5 0

4 years ago