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maks197457 [2]
4 years ago
7

Brent made the table shown to discribe two different relationships between animals.

Physics
1 answer:
grigory [225]4 years ago
5 0

i think it is the second one due to the preditory descritpion of relationship A


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A 77.3 g mass is attached to a horizontal spring with a spring constant of 12.5 N/m and released from rest with an amplitude of
DENIUS [597]

Answer:

speed of the mass is 3.546106 m / s

Explanation:

given data

mass = 77.3 g = 77.3 × 10^{-3}  kg

spring constant k = 12.5 N/m

amplitude A = 38.9 cm = 38.9 ×10^{-2} m

to find out

the speed of the mass

solution

we will apply here conservation energy  that is

K.E + P.E = Total energy  ..................1

so that Total energy = K.E max = P.E max

we know  amplitude so we find out first P.E max that is  

PE max = K.E + P.E  

(1/2)kA² = (1/2)mv² + (1/2)kx²  

kA^² =  mv²+ kx²

so here v²  will be

v²  = k(A² - x²) / m  

v = √[(k/m)×(A² - x²)]  ............2

here x = (1/2)A   so from from 2 equation

v = √[(k/m)×(A² - (A/2)²)]

v = √[(k/m)×(3/4×A²)]

now put all value

v = √[(12.5/ 77.3 × 10^{-3} )×(3/4×(38.9 ×10^{-2})²)]

v = 3.546106 m / s

speed of the mass is 3.546106 m / s

6 0
3 years ago
We wish to obtain a equal sized inverted image of a candel flame on a screen kept at a distance of 4m from the candle flame a) n
soldier1979 [14.2K]

A) To obtain an inverted image of an object of same size a convex lens is used.

B)

In this question we have given

Distance between candle flame and screen=4

it means distance between candle flame and lens +distance between lens and screen =4m

u+v= 4m

u= v

therefore, 2u= 4m

or, u=-2m              ( u is taken negative because it lies on left side of the lens)

and v=2m

we have to find the focal length of convex lens

\frac{1}{f}  =  \frac{1}{v}   -  \frac{1}{u}\frac{1}{f}  =  \frac{1}{2} - ( -  \frac{1}{2} ) \\  \frac{1}{f}  = \frac{1}{1}\\ f = 1m

Hence, Focal length of the lens is 1m

and lens should be placed at a distance of 2m from the flame

4 0
3 years ago
What area must the plates of a capacitor be if they have a charge of 5.7uC and an electric field of 3.1 kV/mm between them? O 0.
kirill [66]

Answer:

Area of the plates of a capacitor, A = 0.208 m²

Explanation:

It is given that,

Charge on the parallel plate capacitor, q = 5.7\ \mu C=5.7\times 10^{-6}\ C

Electric field, E = 3.1 kV/mm = 3100000 V/m

The electric field of a parallel plates capacitor is given by :

E=\dfrac{q}{A\epsilon_o}

A=\dfrac{q}{E\epsilon_o}

A=\dfrac{5.7\times 10^{-6}\ C}{3100000\ V/m\times 8.85\times 10^{-12}\ F/m}

A = 0.208 m²

So, the area of the plates of a capacitor is 0.208 m². Hence, this is the required solution.

8 0
3 years ago
• 500 waves pass by in 2 second. These waves have a wavelength of 6
tamaranim1 [39]

Answer:

6/2times500=1500

Explanation:

s = d/t which means speed equals distance divided by time.

7 0
3 years ago
Which statement best describes beaches?
mylen [45]
A is a good example, i suppose.             























6 0
3 years ago
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