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3 years ago

What are the units for the spring constant, k?

Physics

2 answers:

vitfil [10]3 years ago

5 0

Answer:

C. newtons/meter

Explanation:

To find the units of spring constant you can use the hook's law, which says,

F = k × x where

F = force (unit Newton)

k = spring constant

x = expansion or contraction ( unit m)

So by this equation you can get that

k = F / X

So you get the units as N/m

Tanya [424]3 years ago

4 0

The units for the spring constant, k, are Newtons per meter (N/m).)

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A 4 kg bowling bowl is sitting on a table 1 meter off the ground. How much potential energy does it have?

il63 [147K]

Answer:

$\huge\boxed{\sf P.E. = 39.2\ Joules}$

Explanation:

Given Data:

Mass = m = 4 kg

Acceleration due to gravity = g = 9.8 m/s²

Height = h = 1 m

Required:

Potential Energy = P.E. = ?

Formula:

P.E. = mgh

Solution:

P.E. = (4)(9.8)(1)

P.E. = 39.2 Joules

$\rule[225]{225}{2}$

Hope this helped!

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3 years ago

The ozone layer protects us from the harmful effects of which type of radiation?

leonid [27]

Answer:

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2 years ago

A cello string 0.75 m long has a 220 hz fundamental frequency. find the wave speed along the vibrating string. answer in units o

maxonik [38]

For fundamental frequency of a string to occur, the length of the string has to be half the wavelength. That is,

1/2y = L, where L = length of the string, y = wavelength.

Therefore,
y = 2L = 2*0.75 =1.5 m

Additionally,
y = v/f Where v = wave speed, and f = ferquncy

Then,
v = y*f = 1.5*220 = 330 m/s

4 0

3 years ago

Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

8 0

3 years ago

what is the electric potential at point A in the electric field created by a point charge of 5.5 • 10^-12 C? estimate k as 9.00

Hitman42 [59]

The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v

Explanation:

Given data,

charge = 5.5 x 10¹² C

k =9.00 x 10⁹

The electric potential V of a point charge can found by,

V= kQ / r

Assuming, r=5.00×10⁻² m

V= 5.5 x 10⁻¹²C x 9.00 x 10⁹ / 5.00×10⁻² m

V= 49.5 x 10⁻³/ 5.00×10⁻²

Electric potential V= 0.099 x 10⁻¹v

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2 years ago