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3 years ago

A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a

Physics

1 answer:

Makovka662 [10]3 years ago

5 0

Answer:

0.51 m

Explanation:

Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.

Kinetic energy, KE=½kx²

Where k is spring constant and x is the compression of spring

Potential energy, PE=mgh

Where g is acceleration due to gravity, h is height and m is mass

Equating KE=PE

mgh=½kx²

Making x the subject of formula

$x=\sqrt {\frac {2mgh}{k}}$

Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then

$x=\sqrt \frac {2*1300*9.81*10}{1000000}=0.50503465227646m\\x\approx 0.51 m$

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The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi

Keith_Richards [23]

Answer:

Explanation:

a )

from lens makers formula

$\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})$

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

$\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})$

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens f = 1.8 cm

image distance v = ?

lens formula

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

$\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}$

$\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}$

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

5 0

3 years ago

Scientific evidence documents the pattern of evolution. The evidence exists in a variety of categories. What are these categorie

Sauron [17]

Answer:

Explanation:

Scientific evidence takes note of the pattern of evolution. The evidence exists in a variety of categories, including direct observation of evolutionary change, the fossil record, homology, and biogeography.

Examples of the categories with their respective examples are:

Direct observation of evolutionary change: Development of drug resistant bacteria

Fossil record: Discovery of transitional forms of horses, Discovery of shells of extinct species

Homology: Similarities in mammalian forelimbs, Same genetic code in fireflies and tobacco plants, Vestigial pelvis in right whales

Biogeography: Similarity of endemic island species to nearby mainland species, The high concentration of marsupial species in Australia

3 0

3 years ago

The length of the adult mosquito is typically between 3.0 mm and 6.0 mm. The smallest known mosquitoes are around 2.5 mm, and th

FrozenT [24]

0.74in

Explanation:

Given parameters:

length range of adult mosquito = 3.0 - 6.0mm

Smallest mosquito = 2.5mm

Largest mosquito = 19mm

Average mass range = 3.0 - 5.0mg

Unknown:

Length of largest known mosquito in inches = ?

Solution;

The length is the longest dimension. It is how long a body is.

The problem here is converting from mm to inches;

The length of the longest mosquito which is the largest is 19mm

19mm to inches;

1mm = 0.039inches

19mm = 19mm x $\frac{0.039in}{1mm}$ = 0.74in

learn more:

Scale brainly.com/question/570757

#learnwithBrainly

4 0

3 years ago

The fuzzy cloud around the nucleus is called the

Kamila [148]

It's number three on your worksheet. ;)

6 0

3 years ago

A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats

ololo11 [35]

Answer:

Part a)

$f_B = 290 Hz$

Part B)

percentage increase is

$percentage = 1.38$ %

Explanation:

Part a)

As we know that the beat frequency is

$f_A - f_B = 3$

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

$293 - f_B = 3$

$f_B = 290 Hz$

Part B)

percentage increase in the tension of the string will be given as

$f_A - f_B' = 1$

$f_B' = 292 Hz$

now we have

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

so we have

$T_1 = C (290)^2$

$T_2 = C(292)^2$

so we have

$\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}$

percentage increase is

$percentage = 1.38$

4 0

3 years ago