An action potential in a particular cell has the same amplitude. True or False

In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 27

Vera_Pavlovna [14]

Answer:

9.74 x 10^7 m/s

Explanation:

V = 27000 V

energy of electrons = e x V

K = 1.6 x 10^-19 x 27000 = 43200 x 10^-19 J

Energy = 1/2 m v^2

43200 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v^2 = 9.495 x 10^15

v = 9.74 x 10^7 m/s

4 0

3 years ago

In one hour Earth receives enough solar energy to meet the energy needs of the world for

erastova [34]

I have read that it is enough energy to power the needs of civilization for an entire year.

3 0

3 years ago

A 15 m ladder with a mass of 51 kg is leaning against a frictionless wall, which makes an angle of 60 degrees with the horizonta

timofeeve [1]

Answer:

Explanation:

Given that, .

Mass of ladder is 51kg

Then, it weight is

WL = mg = 51 × 9.81 = 500.31N

This weight will act at the midpoint of the ladder

Length of ladder is 15m

The ladder makes an angle 55°C with the horizontal

An object whose mass is 81kg is at 4m from the bottom of the ladder

Then, weight of object

Wo = mg = 81 × 9.81 = 794.61 N

Using newton second law

Check attachment

Ng is normal force on the ground

Ff is the horizontal frictional force

Nw Is the normal force on the wall

ΣFy = 0

Ng = Wo + WL

Ng = 794.61 + 500.31

Ng = 1294.92 N

Also

ΣFx = 0

Ff — Nw = 0

Then,

Ff = Nw

Now taking moment about point A.

Check attachment

using the principle of equilibrium

Sum of clockwise moment equals to sum of anti-clockwise moment

Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object

Clockwise = Anticlockwise

Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15

794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15

1589.22 + 1876.163 = 12.99•Nw

3465.383 = 12.99•Nw

Nw = 3465.383 / 12.99

Nw = 266.77 N

Since, Nw = Ff

Then, Ff = 266.77N

the horizontal force exerted by the ground on the ladder is 266.77 N

8 0

3 years ago

Read 2 more answers

A golf ball is hit horizontally off the edge of a 30 m high cliff and lands a distance of 25 m from the edge of the cliff. What

Ratling [72]

Answer:

V₀y = 0 m/s

t = 2.47 s

V₀ₓ = 61.86 m/s

Vₓ = 61.86 m/s

Explanation:

Since, the ball is hit horizontally, there is no vertical component of velocity at initial point. So, the initial vertical velocity (V₀y) will beL

V₀y = 0 m/s

For the initial vertical velocity of golf ball we consider the vertical motion and apply 2nd equation of motion:

Y = V₀y*t + (0.5)gt²

where,

Y = Height = 30 m

g = 9.8 m/s²

t = time to hit the ground = ?

Therefore,

30 m = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

t² = 30 m/4.9 m/s²

t = √6.122 s²

t = 2.47 s

For initial vertical velocity we analyze the horizontal motion of the ball. We neglect the frictional effects in horizontal motion thus the speed remains uniform. Hence,

V₀ₓ = Xt

where,

V₀ₓ = Initial vertical Velocity = ?

X = Horizontal Distance = 25 m

Therefore,

V₀ₓ = (25 m)(2.47 s)

V₀ₓ = 61.86 m/s

Due, to uniform motion in horizontal direction:

Final Vertical Velocity = Vₓ = V₀ₓ

Vₓ = 61.86 m/s

4 0

3 years ago

A horizontal spring is attached to the wall on one end and to a mass on the other end. The mass can slide freely on a frictionle

Radda [10]

Answer:

a) x=0 %T=0, b) x= A %T=100%, c) x=-A %T=50%

Explanation:

This is a simple harmonic movement exercise, which is explained by the expression

x = A cos (wt + Ф)

where angular velocity is related to frequency and period

w = 2π f = 2π / T

we can write the equation of the oscillation

x = A cos θ

When seeing the two equations they are equivalent, so what happens with the angle will also happen with time

We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching

The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return

a) for the unstretched spring point x = 0

In general, both distance and time are measured from this point, so the percentage of time is zero.

% T = 0

b) for x = A

let's find the angle

cos tea = x / A = 1

therefore the angles tea = 2π rad

when the movement reaches the point of 2π radians it begins to repeat so the period is complete

% T = 100%

c) the point of maximum compression x = -A

let's look for the angles

cos tea = x / A = -1

therefore the angles tea = π rad

at this point the movement is halfway so it should take half the time

% T = 50%

6 0

3 years ago