Does a resistor resist current or voltage

The magnitude of a force is:

lys-0071 [83]

Answer:

c

Explanation:

force is how hard it is pulled or pushed

3 0

3 years ago

A person is diving in a lake in the depth of h = 5.5 m. The density of the water is rho = 1.0 x10^3 kg/m^3. The pressure of the

Cloud [144]

Answer:a) P = Po + rho×h×g

b) P = 5.4 × 10^9 pa

c) F = P/A = (Po + rho×h×g)/A

d) 1.174×10^11N

Explanation: Using the formula

P = Po + rho×h×g

P = 1.0 x 10^5 + 1000 × 5.5 × 9.81

P = 5.4 × 10^9pa

The magnitude of the force exerted by water on the top of the person's head F at the depth h in terms of P

F = P/A = (Po + rho×h×g)/A

Using the above formula

Where A = 0.046m^2

F = P/ A = 5.4×10^9/0.046

F = 1.174×10^11N

3 0

2 years ago

A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

guajiro [1.7K]

Answer:

$f_{e}$ = 1.7 cm

Explanation:

The magnification of the compound microscope is given by the product of the magnification of each lens

M = M₀ $m_{e}$

M = - L/f₀ 25/ $f_{e}$

Where f₀ and $f_{e}$ are the focal lengths of the lens and eyepiece, respectively, all values in centimeters

In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece ( $f_{e}$ )

$f_{e}$ = - L / f₀ 25 / M

Let's calculate

$f_{e}$ = - 16 / 0.6 25 / (-400)

$f_{e}$ = 1.67 cm

The minus sign in the magnification is because the image is inverted.

$f_{e}$ = 1.7 cm

6 0

3 years ago

A hockey player skates with an average speed of 8.25 m/s. The distance the player will travel in 5.60 s is __________ .

OverLord2011 [107]

<h2>Greetings!</h2>

To find this value, you need to remember the speed formula:

3 = 6 / 2

Speed = distance ÷ time

Rearrange to make distance the subject:

Distance = speed * time

Simply plug these values into this:

5.6 * 8.25 = 46.2

<h3>So the player will travel 46.2 metres!</h3>
<h2>Hope this helps!</h2>

7 0

2 years ago

A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear

JulijaS [17]

Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval (−L,0] for some large number L. The voltage V as a function of x on the interval (0,∞) is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element dx′, we have λ=dqdx′.V(x)=1/4πϵ0∫linedq/r=λ/4πϵ0∫−L0dx/x−x′=λ/4πϵ0(ln|x+L|−ln|x|)

5 0

3 years ago