Being overweight or obese increases the risk of:

A nonconducting rod of length L = 8.15 cm has charge –q = -4.23 fC uniformly distributed along its length.(a) What is the linear

vichka [17]

Answer:

a) λ = 5.19 10⁻⁴ C/m , b) E = 1,573 10⁻³ N/C , c) the direction of the field is directed to the bar

Explanation:

a) the linear density defined as the ratio between the charge per unit length

λ = q / l

Let's start by reducing the units to the SI system

L = 8.15 cm (1m / 100cm) = 8.15 10⁻² m

a = 12 cm (1m / 100cm) = 12 10⁻² m

q = -4.23 fC (1 C / 10¹⁵ ft) = -4.23 10⁻¹⁵ C

λ = -4.23 10⁻¹⁵ C / 8.15 10⁻²

λ = 5.19 10⁻⁴ C/m

b) Let's look for the electric field for a point at a distance a from the end of the bar

E = k dq / r²

To simplify the notation, suppose the bar is the x axis. Since the density is constant, we can write it differentially

λ = dq/dx

dq = λ dx

E = k ∫ λ dx / x²

We integrate and evaluate between the lower limits x = a and higher x = L + a. Here we place the test point at the origin of the system

E = k λ (-1 / x)

E = k λ (-1 /(L + a) + 1 /a)

E = k λ (L /a(L + a)

Let's change the density for its value

E = k (q / L) (L / a (L + a)

E = k q 1 /[a(L + a)]

E = 8.99 10⁹ 4.23 10⁻¹⁵ [1 /12 10⁻²(8.15 10⁻² + 12 10⁻²)]

E = 1,573 10⁻³ N/C

c) the direction of the field is directed to the bar, because it has a negative charge

d) now we change the distance a = 50 cm = 0.50 m

Bar

E = 8.99 10⁹ 4.23 10⁻¹⁵ ( 1 /0.5(0.0815 +0.5))

E = 1,308 10⁻⁴ N/C

Charge point

q = -4.23 10⁻¹⁵ C

E = k q / r²

E = 8.99 10⁹ 4.23 10⁻¹⁵ / 0.5²

E = 1.521 10⁻⁴ N/C

7 0

3 years ago

An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft

sergeinik [125]

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

Given :

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( $K_{A}$ = 0) to the point B at distance l = 0.500m where its kinetic energy is ( $K_{B}$ = 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

Required :

(a) We are asked to find the electric potential VB

(b) We want to determine the magnitude and the direction of the electric field E.

Solution

(a) We are given the values for VA, $K_{B}$ and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

$K_{A} +U_{A} =K_{B} +U_{B}$ .........................................(1)

Where $K_{A}$ = 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential. So, equation (1) will be in the form :

0+qVA= $K_{B}$ +qVB (Divide by q)

VA= $K_{B}$ /q + VB (solve for VB)

VB=VA- $K_{B}$ /q .......................................(2)

We get the relation between VB, VA and $K_{B}$ , now we can plug our values for VA, $K_{B}$ and q into equation (2) to get VB

VB=VA- $K_{B}$ /q

=30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

=80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

VA-VB = $\int\limits^1_0 {E} \, dl$ ...................................(a)

$=E\int\limits^1_0 {} \, dl$

VA-VB=E $l$ (solve for E)

E= VA-VB/ $l$ ..................................(3)

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

E= VA-VB/ $l$

=-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

5 0

3 years ago

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

y₁ = v₀₁ t + ½ g t²

y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

t ’= (t-t₀)

y₂ = v₀₂ t ’+ ½ g t’²

y₂ = 0 + ½ g (t-t₀)²

y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

S = y₁ -y₂

S = ½ g t²– ½ g (t-t₀)²

S = ½ g [t² - (t²- 2 t to + to²)]

S = ½ g (2 t t₀ - t₀²)

S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to. The rock y has not left and the distance increases

For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂

3 0

3 years ago

You move a 75-kg box 35 m. This requires a force of 90 N. how much work is done while moving the box?

Luda [366]

W = F*d.

= 90*35 = 3150J.

4 0

2 years ago

A person walking covers 5 m ikn 10 s how fast is the person moving

aniked [119]

Not what I'd call 'fast' at all.

Speed = (distance covered) / (time to cover the distance) .

Speed = (5 meters) / (10 seconds)

Speed = 0.5 meter per second .

That's like about 1.1 mile per hour .

Normal walking speed is considered to be around 1.4 m/s ... about 3.1 mph, or 14 meters in 10 seconds.

I've got a grandson who hasn't even turned 1 yet. He crawls and doesn't walk, but if you only cover 5m in 10s, he'd leave you in the dust pretty quick.

5 0

3 years ago