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3 years ago

A person who is heavier than the standard for the person’s height is _____

Physics

2 answers:

const2013 [10]3 years ago

5 0

Answer:

Explanation:

since of his lack of growing his body dont know what to do with the needed fat so while he's still growing its in his tummy waiting

N76 [4]3 years ago

4 0

Answer:

A. overweight

Explanation:

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A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predete

zmey [24]

Answer:

The diameter of wire should be $4.04 \times 10^{-4}$ m

Explanation:

Given:

Current density $J = 500 \times 10^{4}$ $\frac{A}{m^{2} }$

Current $I = 0.64$ A

From the formula of current density,

$J = \frac{I}{A}$

Where $A =$ area of cylindrical wire = $\pi r^{2}$

$\pi r^{2} = \frac{I}{J}$

$r^{2} = \frac{I}{\pi J }$

$r = \sqrt{\frac{0.64}{3.14 \times 500 \times 10^{4} } }$

$r = 2.02 \times 10^{-4}$ m

For finding the diameter of wire,

$d = 2r$

$d = 4.04 \times 10^{-4}$ m

Therefore, the diameter of wire should be $4.04 \times 10^{-4}$ m

8 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

where the units of x are length and the numbers 2.6 and 5.1 have appropriate units so that U(x) has units of energy. What is the

KiRa [710]

Answer:

x = 1.00486 m

Explanation:

The complete question is:

" The potential energy between two atoms in a particular molecule has the form U(x) =(2.6)/x^8 −(5.1)/x^4 where the units of x are length and the num- bers 2.6 and 5.1 have appropriate units so that U(x) has units of energy. What is the equilibrium separation of the atoms (that is the distance at which the force between the atoms is zero)? "

Solution:

- The correlation between force F and energy U is given as:

F = - dU / dx

F = - d[(2.6)/x^8 −(5.1)/x^4] / dx

F = 20.8 / x^9 - 20.4 / x^5

- The equilibrium separation distance between atoms is given when Force F is zero:

0 = 20.8 / x^9 - 20.4 / x^5

0 = 20.8 - 20.4*x^4

x^4 = 20.8/20.4

x = ( 20.8/20.4 )^0.25

x = 1.00486 m

5 0

3 years ago

Longitudinal waves do not have

soldier1979 [14.2K]

E. Nonsense longitudinal waves have all of these properties

8 0

3 years ago

Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it

stich3 [128]

Answer:

a. 6.69m/s

b. y=4.48m

c. t=1.43secs

Explanation:

Data given, acceleration,a=35m/s^2

distance covered,d=64cm=0.64m,

a. to determine the speed, we use the equation of motion

initial velocity,u=0m/s

if we substitute values we arrive at

$v^{2}=u^{2}+2as\\v^{2}=0+2*35*0.64\\v^{2}=44.8m/s\\v=\sqrt{44.8}\\ v=6.69m/s\\$

b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2

and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.

Hence we can write the equation above again

$v^{2}=u^{2}-2a(y-2.2)\\$

if we substitute values we have

$v^{2}=u^{2}-2a(y-2.2)\\0=6.69^{2}-2*9.81(y-2.2)\\y-2.2=\frac{44.76}{19.62} \\y=2.28+2.2\\y=4.48m$

c. the time it takes to arrive at 1.83m is obtain by using the equation below

$1.83-2.2=6.69t-\frac{1}{2} *9.81t^{2}\\4.9t^{2}-6.69t-0.37\\using \\t= \frac{-b±\sqrt{b^{2}-4ac} }{2a}\\ where \\a=4.9, b=-6.69, c=-0.37$

if we insert the values, we solve for t , hence t=1.43secs

6 0

3 years ago