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aleksandr82 [10.1K]

2 years ago

7

elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extensi

on of 2.0cm

1 answer:

Citrus2011 [14]2 years ago

6 0

Answer:

40g

Explanation:

20g range > 1.0cm

Therefore,

40g range > 2.0cm

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Newton's first law of motion is the law of inertia. What does it state?

vladimir2022 [97]

Answer:

b is the right answer

Explanation:

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3 years ago

Given: G = 6.672 × 10−11 N · m2 /kg2 Io, a satellite of Jupiter, has an orbital period of 1.24 days and an orbital radius of 4.1

Dahasolnce [82]

Answer:

Mass of Jupiter = 4.173×10^15kg

Explanation:

Using Kepler's 3rd law, it states that the orbital period T is related to the distance,r as:

T^2 = GM/4 pi × r^3

Where G = universal gravitational constant

r = radius

M = masd of jupiter

Rearranging the formular to make M the subject of formular

T^2 × 4 pi = G M × r^3

(T^2 × 4 pi) / (G× r^3) = M

(1.24^2 × 4 × 3.142) /(6.672×10^-11)(4.11×10^8)^3

M = 19.32 /6.672×10^-11)(4.11×10^8)^3

M = 19.32 / 4.63 ×10^15

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6 0

3 years ago

A soap bubble has a diameter of 4mm .Cakculate the pressure inside it if the atmospheric pressure outside is 10^5Nm^2 . The surf

Kipish [7]

Look it up it will give the right anwser

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3 years ago

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Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45 m diamete

Readme [11.4K]

Answer:

In m/s^2:

a=11.3778 m/s^2

In units of g:

a=1.161 g

Explanation:

Since the racing greyhounds are capable of rounding corners at very high speed so we are going use the following formula of acceleration for circular paths.

$a=\frac{v^2}{r}$

where:

v is the speed

r is the radius

Now,

$a=\frac{16^2}{45/2}\\ a=11.3778 m/s^2$

In g units:

$a=\frac{11.3778\ g}{9.8}\\ a=1.161\ g$

7 0

3 years ago

Kelli weighs 425 N, and she is sitting on a playground swing that hangs 0.36 m above the ground. Her mom pulls the swing back an

kodGreya [7K]

Answer:

V = 3.54 m/s

Explanation:

Using the conservation of energy:

$E_i = E_f$

so:

$wh = \frac{1}{2}mV^2$

where w is te weigh of kelly, h the distance that kelly decends, m is the mass of kelly and V the velocity in the lowest position.

So, the mass of kelly is:

m = 425N/9.8 = 43.36 Kg

and h is:

h = 1m-0.36m =0.64m

then, replacing values, we get:

$(425N)(0.64m) = \frac{1}{2}(43.36kg)v^2$

Solving for v:

V = 3.54 m/s

7 0

3 years ago