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Thepotemich [5.8K]
3 years ago
13

If y^2= 3.249 x 10^-11, y = ?

Physics
2 answers:
Leni [432]3 years ago
5 0

Answer:

Scientific Notation: 3.45 x 10^5

E Notation: 3.45e5

maksim [4K]3 years ago
5 0

Answer:

y=5.7×10^-6 probably

Explanation:

to get rid of a square you may take the square root of each side. the square will cancel with the square root to get y=sqrt(3.249×10^-11)

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aliina [53]

Answer: the top one

Explanation: hope this helps

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3 years ago
Suppose that the magnetic field in some region has the form B = kz ˆx (where k is a constant). Find the force on a square loop (
AleksAgata [21]

Answer:

F = ika^2

Explanation:

As we know that loop is placed in YZ plane and magnetic field is along x direction

So here net force on the side of the loop which lies along Y axis is given as

F_1 = i (\vec L \times \vec B)

here we know that on Y axis z = 0

so B = 0

so we have

F_1 = 0

now on the opposite side we have z = a

so magnetic field is given as

B = ka

so force on that side is given as

F = i(\vec L \times \vec B)

F = i(a)(ka) sin90

F_2 = ika^2

so net force on the loop is given as

F = F_1 + F_2

F = ika^2

5 0
3 years ago
An object that is accelerating may be
Elenna [48]

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6 0
2 years ago
How much force must be applied on a blade of length 4cm and thickness of 0.1mm to exert a pressure of 4000000pa?
Viktor [21]

Answer:

F= 403429 kpa

Explanation:

Pressure is the product of force and area

Mathematically,

P=F*A -------where F is force and A is area.

A= 40 *0.1 = 4mm² -----convert to m²

A= 4e⁻⁶ m²

P= 4000000 pa

F= P/A = 4000000/4e⁻⁶

F= 403428793.493 pa

F= 403429 kpa

7 0
3 years ago
A bungee cord can stretch, but it is never compressed. When the distance between the two ends of the cord is less than its unstr
Ksju [112]

Answer:

Explanation:

Given that

g=9.8m/s²

The spring constant is

k=50N/m

The length of the bungee cord is

Lo=32m

Height of bridge which one end of the bungee is tied is 91m

A steel ball of mass 92kg is attached to the other end of the bungee.

The potential energy(Us) of the steel ball before dropped from the bridge is given as

P.E= mgh

P.E= 92×9.8×91

P.E= 82045.6 J

Us= 82045.6 J

Potential energy)(Uc) of the cord is given as

Uc= ½ke²

Where 'e' is the extension

Then the extension is final height extended by cord minus height of cord

e=hf - hi

e=hf - 32

Uc= ½×50×(hf-32)²

Uc=25(hf-32)²

Using conservation of energy,

Then,

The potential energy of free fall equals the potential energy in string

Uc=Us

25(hf-32)²=82045.6

(hf-32)² = 82045.6/25

(hf-32)²=3281.825

Take square root of both sides

√(hf-32)²=√(3281.825)

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hf=89.29m

We neglect the negative sign of the root because the string cannot compressed

3 0
3 years ago
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